neilstrickland.github.io · Special functions The primary special functions are exp, ln, sin, cos, tan, arcsin, arccos, arctan. Things you should know: I The detailed shape of the

Mathematics with Maple

(MAS100)

Introduction

The lecturer is Professor Neil [email protected]

I We will learn how to use Maple, a powerful software package for solvingmathematical problems.

I In the process, we will review and extend many parts of A-levelmathematics, from a new perspective.

Introduction

The lecturer is Professor Neil [email protected] We will learn how to use Maple, a powerful software package for solving

mathematical problems.


Introduction

The lecturer is Professor Neil [email protected] We will learn how to use Maple, a powerful software package for solving

mathematical problems.


Algebraic manipulation

Skills to learn or practice:

I Expand out powers and products

I Factorize simple expressions by inspection

I Manipulate powers (using anam = an+m, (an)m = anm and so on)

I Manipulate and simplify algebraic fractions

Maple commands:

I expand, factor and combine

I simplify; the symbolic option

I collect and coeff







Maple commands:



I collect and coeff







Maple commands:



I collect and coeff







Maple commands:



I collect and coeff







Maple commands:



I collect and coeff







Maple commands:



I collect and coeff







Maple commands:



I collect and coeff







Maple commands:



I collect and coeff







Maple commands:



I collect and coeff







Maple commands:



I collect and coeff

Expansion

I You should practice expanding out products and powers of algebraicexpressions.

I You should check and remember the following identities:

(a + b)(a− b) = a2 − b2

(a + b)2 = a2 + 2ab + b2

(a− b)2 = a2 − 2ab + b2

.

I Often you will need to use these when a and b are themselves complicatedexpressions.

I Example: To simplify (w + x + y + z)2 − (x + y + z)2,put a = w + x + y + z and b = x + y + z . Then

(w + x + y + z)2 − (x + y + z)2 = a2 − b2 = (a + b)(a− b)

= (w + 2x + 2y + 2z)w

= w 2 + 2xw + 2yw + 2zw .

Expansion



(a + b)(a− b) = a2 − b2

(a + b)2 = a2 + 2ab + b2

(a− b)2 = a2 − 2ab + b2

.



(w + x + y + z)2 − (x + y + z)2 = a2 − b2 = (a + b)(a− b)

= (w + 2x + 2y + 2z)w

= w 2 + 2xw + 2yw + 2zw .

Expansion



(a + b)(a− b) = a2 − b2

(a + b)2 = a2 + 2ab + b2

(a− b)2 = a2 − 2ab + b2

.



(w + x + y + z)2 − (x + y + z)2 = a2 − b2 = (a + b)(a− b)

= (w + 2x + 2y + 2z)w

= w 2 + 2xw + 2yw + 2zw .

Expansion


I You should check and remember the following identities:(a + b)(a− b) = a2 − b2

(a + b)2 = a2 + 2ab + b2

(a− b)2 = a2 − 2ab + b2

.



(w + x + y + z)2 − (x + y + z)2 = a2 − b2 = (a + b)(a− b)

= (w + 2x + 2y + 2z)w

= w 2 + 2xw + 2yw + 2zw .

Expansion



(a + b)2 = a2 + 2ab + b2

(a− b)2 = a2 − 2ab + b2

.



(w + x + y + z)2 − (x + y + z)2 = a2 − b2 = (a + b)(a− b)

= (w + 2x + 2y + 2z)w

= w 2 + 2xw + 2yw + 2zw .

Expansion



(a + b)2 = a2 + 2ab + b2

(a− b)2 = a2 − 2ab + b2.



(w + x + y + z)2 − (x + y + z)2 = a2 − b2 = (a + b)(a− b)

= (w + 2x + 2y + 2z)w

= w 2 + 2xw + 2yw + 2zw .

Expansion



(a + b)2 = a2 + 2ab + b2

(a− b)2 = a2 − 2ab + b2.



(w + x + y + z)2 − (x + y + z)2 = a2 − b2 = (a + b)(a− b)

= (w + 2x + 2y + 2z)w

= w 2 + 2xw + 2yw + 2zw .

Expansion



(a + b)2 = a2 + 2ab + b2

(a− b)2 = a2 − 2ab + b2.



(w + x + y + z)2 − (x + y + z)2 = a2 − b2 = (a + b)(a− b)

= (w + 2x + 2y + 2z)w

= w 2 + 2xw + 2yw + 2zw .

Expansion



(a + b)2 = a2 + 2ab + b2

(a− b)2 = a2 − 2ab + b2.



(w + x + y + z)2 − (x + y + z)2 = a2 − b2

= (a + b)(a− b)

= (w + 2x + 2y + 2z)w

= w 2 + 2xw + 2yw + 2zw .

Expansion



(a + b)2 = a2 + 2ab + b2

(a− b)2 = a2 − 2ab + b2.



(w + x + y + z)2 − (x + y + z)2 = a2 − b2 = (a + b)(a− b)

= (w + 2x + 2y + 2z)w

= w 2 + 2xw + 2yw + 2zw .

Expansion



(a + b)2 = a2 + 2ab + b2

(a− b)2 = a2 − 2ab + b2.



(w + x + y + z)2 − (x + y + z)2 = a2 − b2 = (a + b)(a− b)

= (w + 2x + 2y + 2z)w

= w 2 + 2xw + 2yw + 2zw .

Expansion



(a + b)2 = a2 + 2ab + b2

(a− b)2 = a2 − 2ab + b2.



(w + x + y + z)2 − (x + y + z)2 = a2 − b2 = (a + b)(a− b)

= (w + 2x + 2y + 2z)w

= w 2 + 2xw + 2yw + 2zw .

An example: Cauchy-Schwartz

I Problem: Check the identity

(x2 + y 2 + z2)(u2 + v 2 + w 2) = (xu + yv + zw)2 +

(xv − yu)2 + (yw − zv)2 + (zu − xw)2

≥ (xu + yv + zw)2

I .

I . (xu + yv + zw)2 =

+ (xv − yu)2 + x2v 2 − 2xyuv + y 2u2

+ (yw − zv)2 + y 2w 2 − 2yzvw + z2v 2

+ (zu − xw)2 + z2u2 − 2xzuw + x2w 2



(x2 + y 2 + z2)(u2 + v 2 + w 2) = (xu + yv + zw)2 +

(xv − yu)2 + (yw − zv)2 + (zu − xw)2

≥ (xu + yv + zw)2

I .

I . (xu + yv + zw)2 =

+ (xv − yu)2 + x2v 2 − 2xyuv + y 2u2

+ (yw − zv)2 + y 2w 2 − 2yzvw + z2v 2

+ (zu − xw)2 + z2u2 − 2xzuw + x2w 2



(x2 + y 2 + z2)(u2 + v 2 + w 2) = (xu + yv + zw)2 +

(xv − yu)2 + (yw − zv)2 + (zu − xw)2

≥ (xu + yv + zw)2

I .(x2 + y 2 + z2)(u2 + v 2 + w 2)

I . (xu + yv + zw)2 =

+ (xv − yu)2 + x2v 2 − 2xyuv + y 2u2

+ (yw − zv)2 + y 2w 2 − 2yzvw + z2v 2

+ (zu − xw)2 + z2u2 − 2xzuw + x2w 2



(x2 + y 2 + z2)(u2 + v 2 + w 2) = (xu + yv + zw)2 +

(xv − yu)2 + (yw − zv)2 + (zu − xw)2

≥ (xu + yv + zw)2

I .(x2 + y 2 + z2)(u2 + v 2 + w 2) = x2u2 + · · ·

I . (xu + yv + zw)2 =

+ (xv − yu)2 + x2v 2 − 2xyuv + y 2u2

+ (yw − zv)2 + y 2w 2 − 2yzvw + z2v 2

+ (zu − xw)2 + z2u2 − 2xzuw + x2w 2



(x2 + y 2 + z2)(u2 + v 2 + w 2) = (xu + yv + zw)2 +

(xv − yu)2 + (yw − zv)2 + (zu − xw)2

≥ (xu + yv + zw)2

I .(x2 + y 2 + z2)(u2 + v 2 + w 2) = x2u2 + x2v 2 + · · ·

I . (xu + yv + zw)2 =

+ (xv − yu)2 + x2v 2 − 2xyuv + y 2u2

+ (yw − zv)2 + y 2w 2 − 2yzvw + z2v 2

+ (zu − xw)2 + z2u2 − 2xzuw + x2w 2



(x2 + y 2 + z2)(u2 + v 2 + w 2) = (xu + yv + zw)2 +

(xv − yu)2 + (yw − zv)2 + (zu − xw)2

≥ (xu + yv + zw)2

I .(x2 + y 2 + z2)(u2 + v 2 + w 2) = x2u2 + x2v 2 + x2w 2 + · · ·

I . (xu + yv + zw)2 =

+ (xv − yu)2 + x2v 2 − 2xyuv + y 2u2

+ (yw − zv)2 + y 2w 2 − 2yzvw + z2v 2

+ (zu − xw)2 + z2u2 − 2xzuw + x2w 2



(x2 + y 2 + z2)(u2 + v 2 + w 2) = (xu + yv + zw)2 +

(xv − yu)2 + (yw − zv)2 + (zu − xw)2

≥ (xu + yv + zw)2

I .(x2 + y 2 + z2)(u2 + v 2 + w 2) = x2u2 + x2v 2 + x2w 2 +

y 2u2 + · · ·

I . (xu + yv + zw)2 =

+ (xv − yu)2 + x2v 2 − 2xyuv + y 2u2

+ (yw − zv)2 + y 2w 2 − 2yzvw + z2v 2

+ (zu − xw)2 + z2u2 − 2xzuw + x2w 2



(x2 + y 2 + z2)(u2 + v 2 + w 2) = (xu + yv + zw)2 +

(xv − yu)2 + (yw − zv)2 + (zu − xw)2

≥ (xu + yv + zw)2

I .(x2 + y 2 + z2)(u2 + v 2 + w 2) = x2u2 + x2v 2 + x2w 2 +

y 2u2 + y 2v 2 + · · ·

I . (xu + yv + zw)2 =

+ (xv − yu)2 + x2v 2 − 2xyuv + y 2u2

+ (yw − zv)2 + y 2w 2 − 2yzvw + z2v 2

+ (zu − xw)2 + z2u2 − 2xzuw + x2w 2



(x2 + y 2 + z2)(u2 + v 2 + w 2) = (xu + yv + zw)2 +

(xv − yu)2 + (yw − zv)2 + (zu − xw)2

≥ (xu + yv + zw)2

I .(x2 + y 2 + z2)(u2 + v 2 + w 2) = x2u2 + x2v 2 + x2w 2 +

y 2u2 + y 2v 2 + y 2w 2 + · · ·

I . (xu + yv + zw)2 =

+ (xv − yu)2 + x2v 2 − 2xyuv + y 2u2

+ (yw − zv)2 + y 2w 2 − 2yzvw + z2v 2

+ (zu − xw)2 + z2u2 − 2xzuw + x2w 2



(x2 + y 2 + z2)(u2 + v 2 + w 2) = (xu + yv + zw)2 +

(xv − yu)2 + (yw − zv)2 + (zu − xw)2

≥ (xu + yv + zw)2

I .(x2 + y 2 + z2)(u2 + v 2 + w 2) = x2u2 + x2v 2 + x2w 2 +

y 2u2 + y 2v 2 + y 2w 2 +

z2u2 + · · ·

I . (xu + yv + zw)2 =

+ (xv − yu)2 + x2v 2 − 2xyuv + y 2u2

+ (yw − zv)2 + y 2w 2 − 2yzvw + z2v 2

+ (zu − xw)2 + z2u2 − 2xzuw + x2w 2



(x2 + y 2 + z2)(u2 + v 2 + w 2) = (xu + yv + zw)2 +

(xv − yu)2 + (yw − zv)2 + (zu − xw)2

≥ (xu + yv + zw)2

I .(x2 + y 2 + z2)(u2 + v 2 + w 2) = x2u2 + x2v 2 + x2w 2 +

y 2u2 + y 2v 2 + y 2w 2 +

z2u2 + z2v 2 + · · ·

I . (xu + yv + zw)2 =

+ (xv − yu)2 + x2v 2 − 2xyuv + y 2u2

+ (yw − zv)2 + y 2w 2 − 2yzvw + z2v 2

+ (zu − xw)2 + z2u2 − 2xzuw + x2w 2



(x2 + y 2 + z2)(u2 + v 2 + w 2) = (xu + yv + zw)2 +

(xv − yu)2 + (yw − zv)2 + (zu − xw)2

≥ (xu + yv + zw)2

I .(x2 + y 2 + z2)(u2 + v 2 + w 2) = x2u2 + x2v 2 + x2w 2 +

y 2u2 + y 2v 2 + y 2w 2 +

z2u2 + z2v 2 + z2w 2

I . (xu + yv + zw)2 =

+ (xv − yu)2 + x2v 2 − 2xyuv + y 2u2

+ (yw − zv)2 + y 2w 2 − 2yzvw + z2v 2

+ (zu − xw)2 + z2u2 − 2xzuw + x2w 2



(x2 + y 2 + z2)(u2 + v 2 + w 2) = (xu + yv + zw)2 +

(xv − yu)2 + (yw − zv)2 + (zu − xw)2

≥ (xu + yv + zw)2

I .(x2 + y 2 + z2)(u2 + v 2 + w 2) = x2u2 + x2v 2 + x2w 2 +

y 2u2 + y 2v 2 + y 2w 2 +

z2u2 + z2v 2 + z2w 2

I . (xu + yv + zw)2 =

+ (xv − yu)2 + x2v 2 − 2xyuv + y 2u2

+ (yw − zv)2 + y 2w 2 − 2yzvw + z2v 2

+ (zu − xw)2 + z2u2 − 2xzuw + x2w 2



(x2 + y 2 + z2)(u2 + v 2 + w 2) = (xu + yv + zw)2 +

(xv − yu)2 + (yw − zv)2 + (zu − xw)2

≥ (xu + yv + zw)2

I .(x2 + y 2 + z2)(u2 + v 2 + w 2) = x2u2 + x2v 2 + x2w 2 +

y 2u2 + y 2v 2 + y 2w 2 +

z2u2 + z2v 2 + z2w 2

I . (xu + yv + zw)2 = x2u2 + y 2v 2 + z2w 2 + 2xyuv + 2xzuw + 2yzvw

+ (xv − yu)2 + x2v 2 − 2xyuv + y 2u2

+ (yw − zv)2 + y 2w 2 − 2yzvw + z2v 2

+ (zu − xw)2 + z2u2 − 2xzuw + x2w 2



(x2 + y 2 + z2)(u2 + v 2 + w 2) = (xu + yv + zw)2 +

(xv − yu)2 + (yw − zv)2 + (zu − xw)2

≥ (xu + yv + zw)2

I .(x2 + y 2 + z2)(u2 + v 2 + w 2) = x2u2 + x2v 2 + x2w 2 +

y 2u2 + y 2v 2 + y 2w 2 +

z2u2 + z2v 2 + z2w 2


+ (xv − yu)2 + x2v 2 − 2xyuv + y 2u2

+ (yw − zv)2 + y 2w 2 − 2yzvw + z2v 2

+ (zu − xw)2 + z2u2 − 2xzuw + x2w 2



(x2 + y 2 + z2)(u2 + v 2 + w 2) = (xu + yv + zw)2 +

(xv − yu)2 + (yw − zv)2 + (zu − xw)2

≥ (xu + yv + zw)2

I .(x2 + y 2 + z2)(u2 + v 2 + w 2) = x2u2 + x2v 2 + x2w 2 +

y 2u2 + y 2v 2 + y 2w 2 +

z2u2 + z2v 2 + z2w 2


+ (xv − yu)2 + x2v 2 − 2xyuv + y 2u2

+ (yw − zv)2 + y 2w 2 − 2yzvw + z2v 2

+ (zu − xw)2 + z2u2 − 2xzuw + x2w 2



(x2 + y 2 + z2)(u2 + v 2 + w 2) = (xu + yv + zw)2 +

(xv − yu)2 + (yw − zv)2 + (zu − xw)2

≥ (xu + yv + zw)2

I .(x2 + y 2 + z2)(u2 + v 2 + w 2) = x2u2 + x2v 2 + x2w 2 +

y 2u2 + y 2v 2 + y 2w 2 +

z2u2 + z2v 2 + z2w 2


+ (xv − yu)2 + x2v 2 − 2xyuv + y 2u2

+ (yw − zv)2 + y 2w 2 − 2yzvw + z2v 2

+ (zu − xw)2 + z2u2 − 2xzuw + x2w 2



(x2 + y 2 + z2)(u2 + v 2 + w 2) = (xu + yv + zw)2 +

(xv − yu)2 + (yw − zv)2 + (zu − xw)2

≥ (xu + yv + zw)2

I .(x2 + y 2 + z2)(u2 + v 2 + w 2) = x2u2 + x2v 2 + x2w 2 +

y 2u2 + y 2v 2 + y 2w 2 +

z2u2 + z2v 2 + z2w 2


+ (xv − yu)2 + x2v 2 − 2xyuv + y 2u2

+ (yw − zv)2 + y 2w 2 − 2yzvw + z2v 2

+ (zu − xw)2 + z2u2 − 2xzuw + x2w 2

Factoring

I You should practice finding simple factorizations by inspection.

I .

a2 − b2 = (a + b)(a− b)

a3 − b3 = (a2 + ab + b2)(a− b)

ax2 + bx2 + ay 2 + by 2 = (a + b)(x2 + y 2)

1 + t + t2 + t3 = (1 + t)(1 + t2)

u2 − 5u + 6 = (u − 2)(u − 3)

I Maple’s factor command will handle more complicated cases.

Factoring


I .

a2 − b2 = (a + b)(a− b)

a3 − b3 = (a2 + ab + b2)(a− b)

ax2 + bx2 + ay 2 + by 2 = (a + b)(x2 + y 2)

1 + t + t2 + t3 = (1 + t)(1 + t2)

u2 − 5u + 6 = (u − 2)(u − 3)


Factoring


I . a2 − b2

= (a + b)(a− b)

a3 − b3 = (a2 + ab + b2)(a− b)

ax2 + bx2 + ay 2 + by 2 = (a + b)(x2 + y 2)

1 + t + t2 + t3 = (1 + t)(1 + t2)

u2 − 5u + 6 = (u − 2)(u − 3)


Factoring


I . a2 − b2 = (a + b)(a− b)

a3 − b3 = (a2 + ab + b2)(a− b)

ax2 + bx2 + ay 2 + by 2 = (a + b)(x2 + y 2)

1 + t + t2 + t3 = (1 + t)(1 + t2)

u2 − 5u + 6 = (u − 2)(u − 3)


Factoring


I . a2 − b2 = (a + b)(a− b)

a3 − b3

= (a2 + ab + b2)(a− b)

ax2 + bx2 + ay 2 + by 2 = (a + b)(x2 + y 2)

1 + t + t2 + t3 = (1 + t)(1 + t2)

u2 − 5u + 6 = (u − 2)(u − 3)


Factoring


I . a2 − b2 = (a + b)(a− b)

a3 − b3 = (a2 + ab + b2)(a− b)

ax2 + bx2 + ay 2 + by 2 = (a + b)(x2 + y 2)

1 + t + t2 + t3 = (1 + t)(1 + t2)

u2 − 5u + 6 = (u − 2)(u − 3)


Factoring


I . a2 − b2 = (a + b)(a− b)

a3 − b3 = (a2 + ab + b2)(a− b)

ax2 + bx2 + ay 2 + by 2

= (a + b)(x2 + y 2)

1 + t + t2 + t3 = (1 + t)(1 + t2)

u2 − 5u + 6 = (u − 2)(u − 3)


Factoring


I . a2 − b2 = (a + b)(a− b)

a3 − b3 = (a2 + ab + b2)(a− b)

ax2 + bx2 + ay 2 + by 2 = (a + b)(x2 + y 2)

1 + t + t2 + t3 = (1 + t)(1 + t2)

u2 − 5u + 6 = (u − 2)(u − 3)


Factoring


I . a2 − b2 = (a + b)(a− b)

a3 − b3 = (a2 + ab + b2)(a− b)

ax2 + bx2 + ay 2 + by 2 = (a + b)(x2 + y 2)

1 + t + t2 + t3

= (1 + t)(1 + t2)

u2 − 5u + 6 = (u − 2)(u − 3)


Factoring


I . a2 − b2 = (a + b)(a− b)

a3 − b3 = (a2 + ab + b2)(a− b)

ax2 + bx2 + ay 2 + by 2 = (a + b)(x2 + y 2)

1 + t + t2 + t3 = (1 + t)(1 + t2)

u2 − 5u + 6 = (u − 2)(u − 3)


Factoring


I . a2 − b2 = (a + b)(a− b)

a3 − b3 = (a2 + ab + b2)(a− b)

ax2 + bx2 + ay 2 + by 2 = (a + b)(x2 + y 2)

1 + t + t2 + t3 = (1 + t)(1 + t2)

u2 − 5u + 6

= (u − 2)(u − 3)


Factoring


I . a2 − b2 = (a + b)(a− b)

a3 − b3 = (a2 + ab + b2)(a− b)

ax2 + bx2 + ay 2 + by 2 = (a + b)(x2 + y 2)

1 + t + t2 + t3 = (1 + t)(1 + t2)

u2 − 5u + 6 = (u − 2)(u − 3)


Factoring


I . a2 − b2 = (a + b)(a− b)

a3 − b3 = (a2 + ab + b2)(a− b)

ax2 + bx2 + ay 2 + by 2 = (a + b)(x2 + y 2)

1 + t + t2 + t3 = (1 + t)(1 + t2)

u2 − 5u + 6 = (u − 2)(u − 3)


Powers

I You should practice using the basic rules for powers:

anam = an+m (an)m = anm

anbn = (ab)n an/bn = (a/b)n = anb−n

(a + b)n 6=an + bn (a + b)n =∑n

k=0n!

k!(n−k)!akbn−k

I Warning: the rule (an)m = anm has exceptions, for example:

((−3)4)14 = (81)

14 = +3 but (−3)4× 1

4 = (−3)1 = −3.

However, the rule works whenever a > 0 or n and m are integers.

I Example:

(21/231/341/4)3 = 23/233/343/4

= 23/2(22)3/43

= 23/223/23

= 233 = 24

Powers




(a + b)n 6=an + bn (a + b)n =∑n

k=0n!

k!(n−k)!akbn−k


((−3)4)14 = (81)

14 = +3 but (−3)4× 1

4 = (−3)1 = −3.


I Example:

(21/231/341/4)3 = 23/233/343/4

= 23/2(22)3/43

= 23/223/23

= 233 = 24

Powers


anam = an+m

(an)m = anm


(a + b)n 6=an + bn (a + b)n =∑n

k=0n!

k!(n−k)!akbn−k


((−3)4)14 = (81)

14 = +3 but (−3)4× 1

4 = (−3)1 = −3.


I Example:

(21/231/341/4)3 = 23/233/343/4

= 23/2(22)3/43

= 23/223/23

= 233 = 24

Powers




(a + b)n 6=an + bn (a + b)n =∑n

k=0n!

k!(n−k)!akbn−k


((−3)4)14 = (81)

14 = +3 but (−3)4× 1

4 = (−3)1 = −3.


I Example:

(21/231/341/4)3 = 23/233/343/4

= 23/2(22)3/43

= 23/223/23

= 233 = 24

Powers



anbn = (ab)n

an/bn = (a/b)n = anb−n

(a + b)n 6=an + bn (a + b)n =∑n

k=0n!

k!(n−k)!akbn−k


((−3)4)14 = (81)

14 = +3 but (−3)4× 1

4 = (−3)1 = −3.


I Example:

(21/231/341/4)3 = 23/233/343/4

= 23/2(22)3/43

= 23/223/23

= 233 = 24

Powers




(a + b)n 6=an + bn (a + b)n =∑n

k=0n!

k!(n−k)!akbn−k


((−3)4)14 = (81)

14 = +3 but (−3)4× 1

4 = (−3)1 = −3.


I Example:

(21/231/341/4)3 = 23/233/343/4

= 23/2(22)3/43

= 23/223/23

= 233 = 24

Powers




(a + b)n 6=an + bn

(a + b)n =∑n

k=0n!

k!(n−k)!akbn−k


((−3)4)14 = (81)

14 = +3 but (−3)4× 1

4 = (−3)1 = −3.


I Example:

(21/231/341/4)3 = 23/233/343/4

= 23/2(22)3/43

= 23/223/23

= 233 = 24

Powers




(a + b)n 6=an + bn (a + b)n =∑n

k=0n!

k!(n−k)!akbn−k


((−3)4)14 = (81)

14 = +3 but (−3)4× 1

4 = (−3)1 = −3.


I Example:

(21/231/341/4)3 = 23/233/343/4

= 23/2(22)3/43

= 23/223/23

= 233 = 24

Powers




(a + b)n 6=an + bn (a + b)n =∑n

k=0n!

k!(n−k)!akbn−k


((−3)4)14 = (81)

14 = +3 but (−3)4× 1

4 = (−3)1 = −3.


I Example:

(21/231/341/4)3 = 23/233/343/4

= 23/2(22)3/43

= 23/223/23

= 233 = 24

Powers




(a + b)n 6=an + bn (a + b)n =∑n

k=0n!

k!(n−k)!akbn−k


((−3)4)14 = (81)

14 = +3 but (−3)4× 1

4 = (−3)1 = −3.


I Example:

(21/231/341/4)3 = 23/233/343/4

= 23/2(22)3/43

= 23/223/23

= 233 = 24

Powers




(a + b)n 6=an + bn (a + b)n =∑n

k=0n!

k!(n−k)!akbn−k


((−3)4)14 = (81)

14 = +3 but (−3)4× 1

4 = (−3)1 = −3.


I Example:

(21/231/341/4)3

= 23/233/343/4

= 23/2(22)3/43

= 23/223/23

= 233 = 24

Powers




(a + b)n 6=an + bn (a + b)n =∑n

k=0n!

k!(n−k)!akbn−k


((−3)4)14 = (81)

14 = +3 but (−3)4× 1

4 = (−3)1 = −3.


I Example:

(21/231/341/4)3 = 23/233/343/4

= 23/2(22)3/43

= 23/223/23

= 233 = 24

Powers




(a + b)n 6=an + bn (a + b)n =∑n

k=0n!

k!(n−k)!akbn−k


((−3)4)14 = (81)

14 = +3 but (−3)4× 1

4 = (−3)1 = −3.


I Example:

(21/231/341/4)3 = 23/233/343/4

= 23/2(22)3/43

= 23/223/23

= 233 = 24

Powers




(a + b)n 6=an + bn (a + b)n =∑n

k=0n!

k!(n−k)!akbn−k


((−3)4)14 = (81)

14 = +3 but (−3)4× 1

4 = (−3)1 = −3.


I Example:

(21/231/341/4)3 = 23/233/343/4

= 23/2(22)3/43

= 23/223/23

= 233 = 24

Powers




(a + b)n 6=an + bn (a + b)n =∑n

k=0n!

k!(n−k)!akbn−k


((−3)4)14 = (81)

14 = +3 but (−3)4× 1

4 = (−3)1 = −3.


I Example:

(21/231/341/4)3 = 23/233/343/4

= 23/2(22)3/43

= 23/223/23

= 233

= 24

Powers




(a + b)n 6=an + bn (a + b)n =∑n

k=0n!

k!(n−k)!akbn−k


((−3)4)14 = (81)

14 = +3 but (−3)4× 1

4 = (−3)1 = −3.


I Example:

(21/231/341/4)3 = 23/233/343/4

= 23/2(22)3/43

= 23/223/23

= 233 = 24

Algebraic fractions

I You should practice manipulating fractions of the form a/b, where a and bare themselves complicated algebraic expressions.

I The rules are as follows:

a

b+

c

d=

ad + bc

bda

b− c

d=

ad − bc

bda

b.c

d=

ac

bda

b/c

d=

ad

bc( ab

)n=

an

bn( ab

)−n

=bn

an

Algebraic fractions



a

b+

c

d=

ad + bc

bda

b− c

d=

ad − bc

bda

b.c

d=

ac

bda

b/c

d=

ad

bc( ab

)n=

an

bn( ab

)−n

=bn

an

Algebraic fractions



a

b+

c

d=

ad + bc

bda

b− c

d=

ad − bc

bda

b.c

d=

ac

bda

b/c

d=

ad

bc( ab

)n=

an

bn( ab

)−n

=bn

an

Algebraic fractions



a

b+

c

d=

ad + bc

bd

a

b− c

d=

ad − bc

bda

b.c

d=

ac

bda

b/c

d=

ad

bc( ab

)n=

an

bn( ab

)−n

=bn

an

Algebraic fractions



a

b+

c

d=

ad + bc

bda

b− c

d=

ad − bc

bd

a

b.c

d=

ac

bda

b/c

d=

ad

bc( ab

)n=

an

bn( ab

)−n

=bn

an

Algebraic fractions



a

b+

c

d=

ad + bc

bda

b− c

d=

ad − bc

bda

b.c

d=

ac

bd

a

b/c

d=

ad

bc( ab

)n=

an

bn( ab

)−n

=bn

an

Algebraic fractions



a

b+

c

d=

ad + bc

bda

b− c

d=

ad − bc

bda

b.c

d=

ac

bda

b/c

d=

ad

bc

( ab

)n=

an

bn( ab

)−n

=bn

an

Algebraic fractions



a

b+

c

d=

ad + bc

bda

b− c

d=

ad − bc

bda

b.c

d=

ac

bda

b/c

d=

ad

bc( ab

)n=

an

bn

( ab

)−n

=bn

an

Algebraic fractions



a

b+

c

d=

ad + bc

bda

b− c

d=

ad − bc

bda

b.c

d=

ac

bda

b/c

d=

ad

bc( ab

)n=

an

bn( ab

)−n

=bn

an

An example: the cross-ratio

I Put χ(a, b, c, d) = (d−a)(c−b)(d−b)(c−a)

.

I Problem: Show that χ(a, b, c, d) = χ(a−1, b−1, c−1, d−1).

I .

χ(1

a,

1

b,

1

c,

1

d) =

(1d− 1

a

) (1c− 1

b

)(1d− 1

b

) (1c− 1

a

)

=a−dad

b−cbc

b−dbd

a−cac

=(a− d)(b − c)/(abcd)

(b − d)(a− c)/(abcd)

=−− (d − a)(c − b)

−− (d − b)(c − a)

=(d − a)(c − b)

(d − b)(c − a)

= χ(a, b, c, d).



.


I .

χ(1

a,

1

b,

1

c,

1

d) =

(1d− 1

a

) (1c− 1

b

)(1d− 1

b

) (1c− 1

a

)

=a−dad

b−cbc

b−dbd

a−cac

=(a− d)(b − c)/(abcd)


=−− (d − a)(c − b)

−− (d − b)(c − a)

=(d − a)(c − b)

(d − b)(c − a)

= χ(a, b, c, d).



.


I .

χ(1

a,

1

b,

1

c,

1

d) =

(1d− 1

a

) (1c− 1

b

)(1d− 1

b

) (1c− 1

a

)

=a−dad

b−cbc

b−dbd

a−cac

=(a− d)(b − c)/(abcd)


=−− (d − a)(c − b)

−− (d − b)(c − a)

=(d − a)(c − b)

(d − b)(c − a)

= χ(a, b, c, d).



.


I .

χ(1

a,

1

b,

1

c,

1

d) =

(1d− 1

a

) (1c− 1

b

)(1d− 1

b

) (1c− 1

a

)

=a−dad

b−cbc

b−dbd

a−cac

=(a− d)(b − c)/(abcd)


=−− (d − a)(c − b)

−− (d − b)(c − a)

=(d − a)(c − b)

(d − b)(c − a)

= χ(a, b, c, d).



.


I .

χ(1

a,

1

b,

1

c,

1

d) =

(1d− 1

a

) (1c− 1

b

)(1d− 1

b

) (1c− 1

a

)=

a−dad

b−cbc

b−dbd

a−cac

=(a− d)(b − c)/(abcd)


=−− (d − a)(c − b)

−− (d − b)(c − a)

=(d − a)(c − b)

(d − b)(c − a)

= χ(a, b, c, d).



.


I .

χ(1

a,

1

b,

1

c,

1

d) =

(1d− 1

a

) (1c− 1

b

)(1d− 1

b

) (1c− 1

a

)=

a−dad

b−cbc

b−dbd

a−cac

=(a− d)(b − c)/(abcd)


=−− (d − a)(c − b)

−− (d − b)(c − a)

=(d − a)(c − b)

(d − b)(c − a)

= χ(a, b, c, d).



.


I .

χ(1

a,

1

b,

1

c,

1

d) =

(1d− 1

a

) (1c− 1

b

)(1d− 1

b

) (1c− 1

a

)=

a−dad

b−cbc

b−dbd

a−cac

=(a− d)(b − c)/(abcd)


=−− (d − a)(c − b)

−− (d − b)(c − a)

=(d − a)(c − b)

(d − b)(c − a)

= χ(a, b, c, d).



.


I .

χ(1

a,

1

b,

1

c,

1

d) =

(1d− 1

a

) (1c− 1

b

)(1d− 1

b

) (1c− 1

a

)=

a−dad

b−cbc

b−dbd

a−cac

=(a− d)(b − c)/(abcd)


=−− (d − a)(c − b)

−− (d − b)(c − a)

=(d − a)(c − b)

(d − b)(c − a)

= χ(a, b, c, d).



.


I .

χ(1

a,

1

b,

1

c,

1

d) =

(1d− 1

a

) (1c− 1

b

)(1d− 1

b

) (1c− 1

a

)=

a−dad

b−cbc

b−dbd

a−cac

=(a− d)(b − c)/(abcd)


=−− (d − a)(c − b)

−− (d − b)(c − a)

=(d − a)(c − b)

(d − b)(c − a)

= χ(a, b, c, d).

Special functions

The primary special functions are

exp, ln, sin, cos, tan, arcsin, arccos, arctan.

Things you should know:

I The detailed shape of the graphs

I Domains, ranges and inverses

I Properties such as sin(x + y) = sin(x) cos(y) + cos(x) sin(y)

I Derivatives and integrals (covered in later lectures).

The secondary special functions are

sec, csc, cot, sinh, cosh, tanh,sech, csch, coth, arcsinh, arccosh, arctanh.

I You should know how these are defined in terms of the primary functions

(for example, sinh(x) = (exp(x)− exp(−x))/2, and sec(x) = 1/ cos(x))

I You should either remember the properties of the secondary functions, orbe able to derive them from the properties of the primary functions

Special functions













Special functions













Special functions













Special functions













Special functions













Special functions













Special functions













Special functions













Special functions










I You should know how these are defined in terms of the primary functions(for example, sinh(x) = (exp(x)− exp(−x))/2, and sec(x) = 1/ cos(x))


Special functions










I You should know how these are defined in terms of the primary functions(for example, sinh(x) = (exp(x)− exp(−x))/2, and sec(x) = 1/ cos(x))


The exponential function

I exp(x) = 1 + x + x2

2!+ x3

3!+ x4

4!+ · · ·

Warning: infinite sums are subtle.

I e = exp(1) = 1 + 1 + 12!

+ 13!

+ · · · ' 2.71828.

I

exp(x + y) = exp(x) exp(y) exp(x − y) = exp(x)/ exp(y)exp(0) = 1 exp(−x) = 1/ exp(x)

exp(nx) = exp(x)n exp(x) = ex

I

1

e

e2

−2 −1 0 1 2


I exp(x) = 1 + x + x2

2!+ x3

3!+ x4

4!+ · · ·


I e = exp(1) = 1 + 1 + 12!

+ 13!

+ · · · ' 2.71828.

I



I

1

e

e2

−2 −1 0 1 2


I exp(x) = 1 + x + x2

2!+ x3

3!+ x4

4!+ · · ·


I e = exp(1) = 1 + 1 + 12!

+ 13!

+ · · · ' 2.71828.

I



I

1

e

e2

−2 −1 0 1 2


I exp(x) = 1 + x + x2

2!+ x3

3!+ x4

4!+ · · ·


I e = exp(1) = 1 + 1 + 12!

+ 13!

+ · · · ' 2.71828.

I



I

1

e

e2

−2 −1 0 1 2


I exp(x) = 1 + x + x2

2!+ x3

3!+ x4

4!+ · · ·


I e = exp(1) = 1 + 1 + 12!

+ 13!

+ · · · ' 2.71828.

I



I

1

e

e2

−2 −1 0 1 2


I exp(x) = 1 + x + x2

2!+ x3

3!+ x4

4!+ · · ·


I e = exp(1) = 1 + 1 + 12!

+ 13!

+ · · · ' 2.71828.

I



I

1

e

e2

−2 −1 0 1 2

The formula exp(x) exp(y) = exp(x + y)

1 x x2

2!x3

3!x4

4!

y 2xy2!

3x2y3!

4x3y4!

5x4y5!

y2

2!3xy2

3!6x2y2

4!10x3y2

5!15x4y2

6!

y3

3!4xy3

4!10x2y3

5!20x3y3

6!35x4y3

7!

y4

4!5xy4

5!15x2y4

6!35x3y4

7!70x4y4

8!


1 x x2

2!x3

3!x4

4!

y xy x2

2!y x3

3!y x4

4!y

y2

2!x y2

2!x2

2!y2

2!x3

3!y2

2!x4

4!y2

2!

y3

3!x y3

3!x2

2!y3

3!x3

3!y3

3!x4

4!y3

3!

y4

4!x y4

4!x2

2!y4

4!x3

3!y4

4!x4

4!y4

4!

1 x x2

2!x3

3!x4

4!

y 2xy2!

3x2y3!

4x3y4!

5x4y5!

y2

2!3xy2

3!6x2y2

4!10x3y2

5!15x4y2

6!

y3

3!4xy3

4!10x2y3

5!20x3y3

6!35x4y3

7!

y4

4!5xy4

5!15x2y4

6!35x3y4

7!70x4y4

8!


1 x x2

2!x3

3!x4

4!

y 2xy2!

3x2y3!

4x3y4!

5x4y5!

y2

2!3xy2

3!6x2y2

4!10x3y2

5!15x4y2

6!

y3

3!4xy3

4!10x2y3

5!20x3y3

6!35x4y3

7!

y4

4!5xy4

5!15x2y4

6!35x3y4

7!70x4y4

8!


1 x x2

2!x3

3!x4

4!

y 2xy2!

3x2y3!

4x3y4!

5x4y5!

y2

2!3xy2

3!6x2y2

4!10x3y2

5!15x4y2

6!

y3

3!4xy3

4!10x2y3

5!20x3y3

6!35x4y3

7!

y4

4!5xy4

5!15x2y4

6!35x3y4

7!70x4y4

8!

The logarithm

I The natural log function ln(y) is the inverse of the exponential.I ln(y) is defined only when y > 0 (unless we use complex numbers).I We have ln(exp(x)) = ln(ex) = x for all x , and exp(ln(y)) = e ln(y) = y

when y > 0

(NOT ln(x) = 1/ exp(x))

.I

ln(xy) = ln(x) + ln(y) ln(x/y) = ln(x)− ln(y)ln(1) = 0 ln(1/y) = − ln(y)

ln(yn) = n ln(y) ln(e) = 1.

I

1 e e2

−2

−1

0

1

2

ln(x)

The logarithm

I The natural log function ln(y) is the inverse of the exponential.

I ln(y) is defined only when y > 0 (unless we use complex numbers).I We have ln(exp(x)) = ln(ex) = x for all x , and exp(ln(y)) = e ln(y) = y

when y > 0


.I



I

1 e e2

−2

−1

0

1

2

ln(x)

The logarithm

I The natural log function ln(y) is the inverse of the exponential.I ln(y) is defined only when y > 0 (unless we use complex numbers).

I We have ln(exp(x)) = ln(ex) = x for all x , and exp(ln(y)) = e ln(y) = ywhen y > 0


.I



I

1 e e2

−2

−1

0

1

2

ln(x)

The logarithm


when y > 0


.

I



I

1 e e2

−2

−1

0

1

2

ln(x)

The logarithm


when y > 0 (NOT ln(x) = 1/ exp(x)).

I



I

1 e e2

−2

−1

0

1

2

ln(x)

The logarithm


when y > 0 (NOT ln(x) = 1/ exp(x)).I



I

1 e e2

−2

−1

0

1

2

ln(x)

The logarithm


when y > 0 (NOT ln(x) = 1/ exp(x)).I



I

1 e e2

−2

−1

0

1

2

ln(x)

Logs to other bases

I loga(y) is the number t such that y = at (defined for a, y > 0).

I . log10(1000) = log10(103) = 3

log2(1024) = log2(210) = 10

log1024(2) = log1024(10241/10) = 1/10

log3(1/9) = log3(3−2) = −2

I loga(y) = ln(y)/ ln(a)

I Check: aln(y)/ ln(a) = (e ln(a))ln(y)/ ln(a) = e ln(y) = y .

I log10(y) = the number t such that 10t = y' the number of digits in y left of the decimal point.

I This is mostly of historical importance.

I log2(y) = the number t such that 2t = y' the number of bits in y .

I This is of some use in computer science and information theory.

I loge(y) = (the number t such that et = y) = ln(y) = log(y).

Logs to other bases


I . log10(1000) = log10(103) = 3

log2(1024) = log2(210) = 10

log1024(2) = log1024(10241/10) = 1/10

log3(1/9) = log3(3−2) = −2








Logs to other bases


I . log10(1000) = log10(103) = 3

log2(1024) = log2(210) = 10

log1024(2) = log1024(10241/10) = 1/10

log3(1/9) = log3(3−2) = −2








Logs to other bases


I . log10(1000) = log10(103) = 3

log2(1024) = log2(210) = 10

log1024(2) = log1024(10241/10) = 1/10

log3(1/9) = log3(3−2) = −2








Logs to other bases


I . log10(1000) = log10(103) = 3

log2(1024) = log2(210) = 10

log1024(2) = log1024(10241/10) = 1/10

log3(1/9) = log3(3−2) = −2








Logs to other bases


I . log10(1000) = log10(103) = 3

log2(1024) = log2(210) = 10

log1024(2) = log1024(10241/10) = 1/10

log3(1/9) = log3(3−2) = −2








Logs to other bases


I . log10(1000) = log10(103) = 3

log2(1024) = log2(210) = 10

log1024(2) = log1024(10241/10) = 1/10

log3(1/9) = log3(3−2) = −2








Logs to other bases


I . log10(1000) = log10(103) = 3

log2(1024) = log2(210) = 10

log1024(2) = log1024(10241/10) = 1/10

log3(1/9) = log3(3−2) = −2








Logs to other bases


I . log10(1000) = log10(103) = 3

log2(1024) = log2(210) = 10

log1024(2) = log1024(10241/10) = 1/10

log3(1/9) = log3(3−2) = −2








Logs to other bases


I . log10(1000) = log10(103) = 3

log2(1024) = log2(210) = 10

log1024(2) = log1024(10241/10) = 1/10

log3(1/9) = log3(3−2) = −2








Hyperbolic functions

I The hyperbolic functions are defined as follows:

sinh(x) = ex−e−x

2tanh(x) = sinh(x)

cosh(x)csch(x) = 1

sinh(x)

cosh(x) = ex+e−x

2coth(x) = cosh(x)

sinh(x)sech(x) = 1

cosh(x)

Use convert(...,exp) in Maple to rewrite in terms of exponentials.

I Properties are easily deduced from those of exp.I These are related to trig functions using complex numbers, eg

sin(x) = sinh(ix)/i , where i =√−1.

I

sinh(x)

cosh(x)

tanh(x)




2tanh(x) = sinh(x)

cosh(x)csch(x) = 1

sinh(x)

cosh(x) = ex+e−x

2coth(x) = cosh(x)

sinh(x)sech(x) = 1

cosh(x)

Use convert(...,exp) in Maple to rewrite in terms of exponentials.I Properties are easily deduced from those of exp.I These are related to trig functions using complex numbers, eg


I

sinh(x)

cosh(x)

tanh(x)




2

tanh(x) = sinh(x)cosh(x)

csch(x) = 1sinh(x)

cosh(x) = ex+e−x

2coth(x) = cosh(x)

sinh(x)sech(x) = 1

cosh(x)



I

sinh(x)

cosh(x)

tanh(x)




2

tanh(x) = sinh(x)cosh(x)

csch(x) = 1sinh(x)

cosh(x) = ex+e−x

2

coth(x) = cosh(x)sinh(x)

sech(x) = 1cosh(x)



I

sinh(x)

cosh(x)

tanh(x)




2tanh(x) = sinh(x)

cosh(x)

csch(x) = 1sinh(x)

cosh(x) = ex+e−x

2

coth(x) = cosh(x)sinh(x)

sech(x) = 1cosh(x)



I

sinh(x)

cosh(x)

tanh(x)




2tanh(x) = sinh(x)

cosh(x)

csch(x) = 1sinh(x)

cosh(x) = ex+e−x

2coth(x) = cosh(x)

sinh(x)

sech(x) = 1cosh(x)



I

sinh(x)

cosh(x)

tanh(x)




2tanh(x) = sinh(x)

cosh(x)

csch(x) = 1sinh(x)

cosh(x) = ex+e−x

2coth(x) = cosh(x)

sinh(x)sech(x) = 1

cosh(x)



I

sinh(x)

cosh(x)

tanh(x)




2tanh(x) = sinh(x)

cosh(x)csch(x) = 1

sinh(x)

cosh(x) = ex+e−x

2coth(x) = cosh(x)

sinh(x)sech(x) = 1

cosh(x)



I

sinh(x)

cosh(x)

tanh(x)




2tanh(x) = sinh(x)

cosh(x)csch(x) = 1

sinh(x)

cosh(x) = ex+e−x

2coth(x) = cosh(x)

sinh(x)sech(x) = 1

cosh(x)

Use convert(...,exp) in Maple to rewrite in terms of exponentials.

I Properties are easily deduced from those of exp.I These are related to trig functions using complex numbers, eg


I

sinh(x)

cosh(x)

tanh(x)




2tanh(x) = sinh(x)

cosh(x)csch(x) = 1

sinh(x)

cosh(x) = ex+e−x

2coth(x) = cosh(x)

sinh(x)sech(x) = 1

cosh(x)

Use convert(...,exp) in Maple to rewrite in terms of exponentials.I Properties are easily deduced from those of exp.

I These are related to trig functions using complex numbers, egsin(x) = sinh(ix)/i , where i =

√−1.

I

sinh(x)

cosh(x)

tanh(x)




2tanh(x) = sinh(x)

cosh(x)csch(x) = 1

sinh(x)

cosh(x) = ex+e−x

2coth(x) = cosh(x)

sinh(x)sech(x) = 1

cosh(x)



I

sinh(x)

cosh(x)

tanh(x)




2tanh(x) = sinh(x)

cosh(x)csch(x) = 1

sinh(x)

cosh(x) = ex+e−x

2coth(x) = cosh(x)

sinh(x)sech(x) = 1

cosh(x)



I

sinh(x)

cosh(x)

tanh(x)

Hyperbolic identities

I .

cosh(x)2 − sinh(x)2 = 1

sech(x)2 + tanh(x)2 = 1

sinh(x + y) = sinh(x) cosh(y) + cosh(x) sinh(y)

cosh(x + y) = cosh(x) cosh(y) + sinh(x) sinh(y)

I To check these, put u = ex , so sinh(x) = u−u−1

2and cosh(x) = u+u−1

2.

I .

cosh(x)2 − sinh(x)2 =(u + u−1)2

4− (u − u−1)2

4

=(u2 + 2 + u−2)− (u2 − 2 + u−2)

4

= (2− (−2))/4 = 1.

I Now put v = ey , so uv = ex+y .

I sinh(x) cosh(y) + cosh(x) sinh(y)

= (u−u−1)2

(v+v−1)2

+ (u+u−1)2

(v−v−1)2

=(uv + uv−1 − u−1v − u−1v−1 + uv − uv−1 + u−1v − u−1v−1)

4

=uv − (uv)−1

2=

ex+y − e−x−y

2= sinh(x + y)


I . cosh(x)2 − sinh(x)2 = 1






2.

I .

cosh(x)2 − sinh(x)2 =(u + u−1)2

4− (u − u−1)2

4

=(u2 + 2 + u−2)− (u2 − 2 + u−2)

4

= (2− (−2))/4 = 1.



= (u−u−1)2

(v+v−1)2

+ (u+u−1)2

(v−v−1)2


4

=uv − (uv)−1

2=

ex+y − e−x−y

2= sinh(x + y)








2.

I .

cosh(x)2 − sinh(x)2 =(u + u−1)2

4− (u − u−1)2

4

=(u2 + 2 + u−2)− (u2 − 2 + u−2)

4

= (2− (−2))/4 = 1.



= (u−u−1)2

(v+v−1)2

+ (u+u−1)2

(v−v−1)2


4

=uv − (uv)−1

2=

ex+y − e−x−y

2= sinh(x + y)








2.

I .

cosh(x)2 − sinh(x)2 =(u + u−1)2

4− (u − u−1)2

4

=(u2 + 2 + u−2)− (u2 − 2 + u−2)

4

= (2− (−2))/4 = 1.



= (u−u−1)2

(v+v−1)2

+ (u+u−1)2

(v−v−1)2


4

=uv − (uv)−1

2=

ex+y − e−x−y

2= sinh(x + y)








2.

I .

cosh(x)2 − sinh(x)2 =(u + u−1)2

4− (u − u−1)2

4

=(u2 + 2 + u−2)− (u2 − 2 + u−2)

4

= (2− (−2))/4 = 1.



= (u−u−1)2

(v+v−1)2

+ (u+u−1)2

(v−v−1)2


4

=uv − (uv)−1

2=

ex+y − e−x−y

2= sinh(x + y)








2.

I .

cosh(x)2 − sinh(x)2 =(u + u−1)2

4− (u − u−1)2

4

=(u2 + 2 + u−2)− (u2 − 2 + u−2)

4

= (2− (−2))/4 = 1.



= (u−u−1)2

(v+v−1)2

+ (u+u−1)2

(v−v−1)2


4

=uv − (uv)−1

2=

ex+y − e−x−y

2= sinh(x + y)








2.

I .

cosh(x)2 − sinh(x)2

=(u + u−1)2

4− (u − u−1)2

4

=(u2 + 2 + u−2)− (u2 − 2 + u−2)

4

= (2− (−2))/4 = 1.



= (u−u−1)2

(v+v−1)2

+ (u+u−1)2

(v−v−1)2


4

=uv − (uv)−1

2=

ex+y − e−x−y

2= sinh(x + y)








2.

I .

cosh(x)2 − sinh(x)2 =(u + u−1)2

4− (u − u−1)2

4

=(u2 + 2 + u−2)− (u2 − 2 + u−2)

4

= (2− (−2))/4 = 1.



= (u−u−1)2

(v+v−1)2

+ (u+u−1)2

(v−v−1)2


4

=uv − (uv)−1

2=

ex+y − e−x−y

2= sinh(x + y)








2.

I .

cosh(x)2 − sinh(x)2 =(u + u−1)2

4− (u − u−1)2

4

=(u2 + 2 + u−2)− (u2 − 2 + u−2)

4

= (2− (−2))/4 = 1.



= (u−u−1)2

(v+v−1)2

+ (u+u−1)2

(v−v−1)2


4

=uv − (uv)−1

2=

ex+y − e−x−y

2= sinh(x + y)








2.

I .

cosh(x)2 − sinh(x)2 =(u + u−1)2

4− (u − u−1)2

4

=(u2 + 2 + u−2)− (u2 − 2 + u−2)

4

= (2− (−2))/4 = 1.



= (u−u−1)2

(v+v−1)2

+ (u+u−1)2

(v−v−1)2


4

=uv − (uv)−1

2=

ex+y − e−x−y

2= sinh(x + y)








2.

I .

cosh(x)2 − sinh(x)2 =(u + u−1)2

4− (u − u−1)2

4

=(u2 + 2 + u−2)− (u2 − 2 + u−2)

4

= (2− (−2))/4 = 1.



= (u−u−1)2

(v+v−1)2

+ (u+u−1)2

(v−v−1)2


4

=uv − (uv)−1

2=

ex+y − e−x−y

2= sinh(x + y)








2.

I .

cosh(x)2 − sinh(x)2 =(u + u−1)2

4− (u − u−1)2

4

=(u2 + 2 + u−2)− (u2 − 2 + u−2)

4

= (2− (−2))/4 = 1.



= (u−u−1)2

(v+v−1)2

+ (u+u−1)2

(v−v−1)2


4

=uv − (uv)−1

2=

ex+y − e−x−y

2= sinh(x + y)








2.

I .

cosh(x)2 − sinh(x)2 =(u + u−1)2

4− (u − u−1)2

4

=(u2 + 2 + u−2)− (u2 − 2 + u−2)

4

= (2− (−2))/4 = 1.


I sinh(x) cosh(y) + cosh(x) sinh(y) = (u−u−1)2

(v+v−1)2

+ (u+u−1)2

(v−v−1)2


4

=uv − (uv)−1

2=

ex+y − e−x−y

2= sinh(x + y)








2.

I .

cosh(x)2 − sinh(x)2 =(u + u−1)2

4− (u − u−1)2

4

=(u2 + 2 + u−2)− (u2 − 2 + u−2)

4

= (2− (−2))/4 = 1.



(v+v−1)2

+ (u+u−1)2

(v−v−1)2


4

=uv − (uv)−1

2=

ex+y − e−x−y

2= sinh(x + y)








2.

I .

cosh(x)2 − sinh(x)2 =(u + u−1)2

4− (u − u−1)2

4

=(u2 + 2 + u−2)− (u2 − 2 + u−2)

4

= (2− (−2))/4 = 1.



(v+v−1)2

+ (u+u−1)2

(v−v−1)2


4

=uv − (uv)−1

2

=ex+y − e−x−y

2= sinh(x + y)








2.

I .

cosh(x)2 − sinh(x)2 =(u + u−1)2

4− (u − u−1)2

4

=(u2 + 2 + u−2)− (u2 − 2 + u−2)

4

= (2− (−2))/4 = 1.



(v+v−1)2

+ (u+u−1)2

(v−v−1)2


4

=uv − (uv)−1

2=

ex+y − e−x−y

2

= sinh(x + y)








2.

I .

cosh(x)2 − sinh(x)2 =(u + u−1)2

4− (u − u−1)2

4

=(u2 + 2 + u−2)− (u2 − 2 + u−2)

4

= (2− (−2))/4 = 1.



(v+v−1)2

+ (u+u−1)2

(v−v−1)2


4

=uv − (uv)−1

2=

ex+y − e−x−y

2= sinh(x + y)

Inverse hyperbolic functions

I The graph of y = sinh(x) crosses each hori-zontal line precisely once, which means thatthere is an inverse function x = sinh−1(y) =arcsinh(y), defined for all y ∈ R.

I This can be written in terms of ln: arcsinh(y) = ln(y +√

1 + y 2).

I Check: Suppose y = sinh(x); we must show that x = ln(y +√

1 + y 2).

I We have 1 + y2 = 1 + sinh(x)2 = cosh(x)2 (and cosh(x), 1 + y2 > 0), so√1 + y2 = cosh(x).

I Thus y +√

1 + y2 = sinh(x) + cosh(x) = ex−e−x

2+ ex+e−x

2= ex

I so ln(y +√

1 + y2) = ln(ex ) = x as required.

I Similarly, arccosh(y) = ln(y +√

y 2 − 1), defined for y ≥ 1

I and arctanh(y) = 12

ln(

1+y1−y

), defined when −1 < y < 1.




1 + y 2).


1 + y 2).


I Thus y +√


2+ ex+e−x

2= ex

I so ln(y +√





ln(

1+y1−y




x=arcsinh(y)

sinh(x)=y


1 + y 2).


1 + y 2).


I Thus y +√


2+ ex+e−x

2= ex

I so ln(y +√





ln(

1+y1−y




x=arcsinh(y)

sinh(x)=y


1 + y 2).


1 + y 2).


I Thus y +√


2+ ex+e−x

2= ex

I so ln(y +√





ln(

1+y1−y




x=arcsinh(y)

sinh(x)=y


1 + y 2).


1 + y 2).


I Thus y +√


2+ ex+e−x

2= ex

I so ln(y +√





ln(

1+y1−y




x=arcsinh(y)

sinh(x)=y


1 + y 2).


1 + y 2).I We have 1 + y2 = 1 + sinh(x)2 = cosh(x)2 (and cosh(x), 1 + y2 > 0), so√

1 + y2 = cosh(x).

I Thus y +√


2+ ex+e−x

2= ex

I so ln(y +√





ln(

1+y1−y




x=arcsinh(y)

sinh(x)=y


1 + y 2).



1 + y2 = cosh(x).

I Thus y +√


2+ ex+e−x

2= ex

I so ln(y +√





ln(

1+y1−y




x=arcsinh(y)

sinh(x)=y


1 + y 2).



1 + y2 = cosh(x).

I Thus y +√


2+ ex+e−x

2= ex

I so ln(y +√





ln(

1+y1−y




x=arcsinh(y)

sinh(x)=y


1 + y 2).



1 + y2 = cosh(x).

I Thus y +√


2+ ex+e−x

2= ex

I so ln(y +√





ln(

1+y1−y




x=arcsinh(y)

sinh(x)=y


1 + y 2).



1 + y2 = cosh(x).

I Thus y +√


2+ ex+e−x

2= ex

I so ln(y +√





ln(

1+y1−y


Graphs

sinh(x) cosh(x) tanh(x)

arcsinh(x) arccosh(x) arctanh(x)

Trigonometric functions

I Let P be one unit away from the origin, at an angle of θ measuredanticlockwise from the point A = (1, 0).

cos(θ)

sin(θ)

P=(cos(θ),sin(θ))

A=(1,0)

θ

I (We measure θ in radians, so the length of the arc AP is θ.)

I The numbers cos(θ) and sin(θ) are defined to be the x and y coordinatesof P.

I We also put

tan(x) = sin(x)cos(x)

csc(x) = 1sin(x)

cot(x) = cos(x)sin(x)

sec(x) = 1cos(x)



cos(θ)

sin(θ)

P=(cos(θ),sin(θ))

A=(1,0)

θ



I We also put


csc(x) = 1sin(x)


sec(x) = 1cos(x)



cos(θ)

sin(θ)

P=(cos(θ),sin(θ))

A=(1,0)

θ



I We also put


csc(x) = 1sin(x)


sec(x) = 1cos(x)



cos(θ)

sin(θ)

P=(cos(θ),sin(θ))

A=(1,0)

θ



I We also put


csc(x) = 1sin(x)


sec(x) = 1cos(x)



cos(θ)

sin(θ)

P=(cos(θ),sin(θ))

A=(1,0)

θ



I We also puttan(x) = sin(x)

cos(x)

csc(x) = 1sin(x)


sec(x) = 1cos(x)



cos(θ)

sin(θ)

P=(cos(θ),sin(θ))

A=(1,0)

θ




cos(x)csc(x) = 1

sin(x)


sec(x) = 1cos(x)



cos(θ)

sin(θ)

P=(cos(θ),sin(θ))

A=(1,0)

θ




cos(x)csc(x) = 1

sin(x)


sec(x) = 1cos(x)



cos(θ)

sin(θ)

P=(cos(θ),sin(θ))

A=(1,0)

θ




cos(x)csc(x) = 1

sin(x)


sec(x) = 1cos(x)

Graphs

−4π −2π π 2π 3π 4π

sin(θ)

−4π −2π π 2π 3π 4π

cos(θ)

−2π −π π2 π 3π

2 2π

tan(θ)

−2π −π π2 π 3π

2 2π

cot(θ)

Graphs

−4π −2π π 2π 3π 4π

sin(θ)

−4π −2π π 2π 3π 4π

cos(θ)

−2π −π π2 π 3π

2 2π

tan(θ)

−2π −π π2 π 3π

2 2π

cot(θ)

sin(π/2 + x) = cos(x) cos(π/2 + x) = − sin(x)sin(π + x) = − sin(x) cos(π + x) = − cos(x)

sin(2π + x) = sin(x) cos(2π + x) = cos(x)sin(−x) = − sin(x) cos(−x) = cos(x).

Preview of complex numbers

I Complex numbers are expressions like z = 3 + 4i , where i satisfies i2 = −1.I You can add and subtract complex numbers in an obvious way, for

example (3 + 4i) + (7− 3i) = 10 + i .I To multiply: expand out and use i2 = −1. For example:

(1 + 2i)(3 + 4i)

= 3 + 4i + 6i + 8i2 = 3 + 4i + 6i − 8 = −5 + 10i .

I Note that the powers of i repeat with period 4:

i0 = 1 i1 = i i2 = −1 i3 = −i i4 = 1 i5 = i i6 = −1 i7 = −i i8 = 1.

I By expanding and using this we find powers of any complex number.

(1 + i)2 = 1 + 2i + i2 = 1 + 2i + (−1) = 2i

(1 + i)8 = ((1 + i)2)4 = 24i4 = 24 = 16

I Note that

exp(ix) =1 + ix +(ix)2

2+

(ix)3

6+

(ix)4

24+

(ix)5

120+ · · ·

= 1 + ix −x2

2− i

x3

6+

x4

24+ i

x5

120+ · · ·

=

(1−

x2

2+

x4

24+ · · ·

)+

(x −

x3

6+

x5

120+ · · ·

)i

= cos(x) + sin(x)i .


I Complex numbers are expressions like z = 3 + 4i , where i satisfies i2 = −1.

I You can add and subtract complex numbers in an obvious way, forexample (3 + 4i) + (7− 3i) = 10 + i .

I To multiply: expand out and use i2 = −1. For example:(1 + 2i)(3 + 4i)

= 3 + 4i + 6i + 8i2 = 3 + 4i + 6i − 8 = −5 + 10i .


i0 = 1 i1 = i i2 = −1 i3 = −i i4 = 1 i5 = i i6 = −1 i7 = −i i8 = 1.


(1 + i)2 = 1 + 2i + i2 = 1 + 2i + (−1) = 2i

(1 + i)8 = ((1 + i)2)4 = 24i4 = 24 = 16

I Note that


2+

(ix)3

6+

(ix)4

24+

(ix)5

120+ · · ·

= 1 + ix −x2

2− i

x3

6+

x4

24+ i

x5

120+ · · ·

=

(1−

x2

2+

x4

24+ · · ·

)+

(x −

x3

6+

x5

120+ · · ·

)i




example (3 + 4i) + (7− 3i) = 10 + i .

I To multiply: expand out and use i2 = −1. For example:(1 + 2i)(3 + 4i)

= 3 + 4i + 6i + 8i2 = 3 + 4i + 6i − 8 = −5 + 10i .


i0 = 1 i1 = i i2 = −1 i3 = −i i4 = 1 i5 = i i6 = −1 i7 = −i i8 = 1.


(1 + i)2 = 1 + 2i + i2 = 1 + 2i + (−1) = 2i

(1 + i)8 = ((1 + i)2)4 = 24i4 = 24 = 16

I Note that


2+

(ix)3

6+

(ix)4

24+

(ix)5

120+ · · ·

= 1 + ix −x2

2− i

x3

6+

x4

24+ i

x5

120+ · · ·

=

(1−

x2

2+

x4

24+ · · ·

)+

(x −

x3

6+

x5

120+ · · ·

)i





(1 + 2i)(3 + 4i)

= 3 + 4i + 6i + 8i2 = 3 + 4i + 6i − 8 = −5 + 10i .I Note that the powers of i repeat with period 4:

i0 = 1 i1 = i i2 = −1 i3 = −i i4 = 1 i5 = i i6 = −1 i7 = −i i8 = 1.


(1 + i)2 = 1 + 2i + i2 = 1 + 2i + (−1) = 2i

(1 + i)8 = ((1 + i)2)4 = 24i4 = 24 = 16

I Note that


2+

(ix)3

6+

(ix)4

24+

(ix)5

120+ · · ·

= 1 + ix −x2

2− i

x3

6+

x4

24+ i

x5

120+ · · ·

=

(1−

x2

2+

x4

24+ · · ·

)+

(x −

x3

6+

x5

120+ · · ·

)i





(1 + 2i)(3 + 4i) = 3 + 4i + 6i + 8i2

= 3 + 4i + 6i − 8 = −5 + 10i .I Note that the powers of i repeat with period 4:

i0 = 1 i1 = i i2 = −1 i3 = −i i4 = 1 i5 = i i6 = −1 i7 = −i i8 = 1.


(1 + i)2 = 1 + 2i + i2 = 1 + 2i + (−1) = 2i

(1 + i)8 = ((1 + i)2)4 = 24i4 = 24 = 16

I Note that


2+

(ix)3

6+

(ix)4

24+

(ix)5

120+ · · ·

= 1 + ix −x2

2− i

x3

6+

x4

24+ i

x5

120+ · · ·

=

(1−

x2

2+

x4

24+ · · ·

)+

(x −

x3

6+

x5

120+ · · ·

)i





(1 + 2i)(3 + 4i) = 3 + 4i + 6i + 8i2 = 3 + 4i + 6i − 8

= −5 + 10i .I Note that the powers of i repeat with period 4:

i0 = 1 i1 = i i2 = −1 i3 = −i i4 = 1 i5 = i i6 = −1 i7 = −i i8 = 1.


(1 + i)2 = 1 + 2i + i2 = 1 + 2i + (−1) = 2i

(1 + i)8 = ((1 + i)2)4 = 24i4 = 24 = 16

I Note that


2+

(ix)3

6+

(ix)4

24+

(ix)5

120+ · · ·

= 1 + ix −x2

2− i

x3

6+

x4

24+ i

x5

120+ · · ·

=

(1−

x2

2+

x4

24+ · · ·

)+

(x −

x3

6+

x5

120+ · · ·

)i





(1 + 2i)(3 + 4i) = 3 + 4i + 6i + 8i2 = 3 + 4i + 6i − 8 = −5 + 10i .


i0 = 1 i1 = i i2 = −1 i3 = −i i4 = 1 i5 = i i6 = −1 i7 = −i i8 = 1.


(1 + i)2 = 1 + 2i + i2 = 1 + 2i + (−1) = 2i

(1 + i)8 = ((1 + i)2)4 = 24i4 = 24 = 16

I Note that


2+

(ix)3

6+

(ix)4

24+

(ix)5

120+ · · ·

= 1 + ix −x2

2− i

x3

6+

x4

24+ i

x5

120+ · · ·

=

(1−

x2

2+

x4

24+ · · ·

)+

(x −

x3

6+

x5

120+ · · ·

)i





(1 + 2i)(3 + 4i) = 3 + 4i + 6i + 8i2 = 3 + 4i + 6i − 8 = −5 + 10i .I Note that the powers of i repeat with period 4:

i0 = 1 i1 = i i2 = −1 i3 = −i i4 = 1 i5 = i i6 = −1 i7 = −i i8 = 1.


(1 + i)2 = 1 + 2i + i2 = 1 + 2i + (−1) = 2i

(1 + i)8 = ((1 + i)2)4 = 24i4 = 24 = 16

I Note that


2+

(ix)3

6+

(ix)4

24+

(ix)5

120+ · · ·

= 1 + ix −x2

2− i

x3

6+

x4

24+ i

x5

120+ · · ·

=

(1−

x2

2+

x4

24+ · · ·

)+

(x −

x3

6+

x5

120+ · · ·

)i






i0 = 1

i1 = i i2 = −1 i3 = −i i4 = 1 i5 = i i6 = −1 i7 = −i i8 = 1.


(1 + i)2 = 1 + 2i + i2 = 1 + 2i + (−1) = 2i

(1 + i)8 = ((1 + i)2)4 = 24i4 = 24 = 16

I Note that


2+

(ix)3

6+

(ix)4

24+

(ix)5

120+ · · ·

= 1 + ix −x2

2− i

x3

6+

x4

24+ i

x5

120+ · · ·

=

(1−

x2

2+

x4

24+ · · ·

)+

(x −

x3

6+

x5

120+ · · ·

)i






i0 = 1 i1 = i

i2 = −1 i3 = −i i4 = 1 i5 = i i6 = −1 i7 = −i i8 = 1.


(1 + i)2 = 1 + 2i + i2 = 1 + 2i + (−1) = 2i

(1 + i)8 = ((1 + i)2)4 = 24i4 = 24 = 16

I Note that


2+

(ix)3

6+

(ix)4

24+

(ix)5

120+ · · ·

= 1 + ix −x2

2− i

x3

6+

x4

24+ i

x5

120+ · · ·

=

(1−

x2

2+

x4

24+ · · ·

)+

(x −

x3

6+

x5

120+ · · ·

)i






i0 = 1 i1 = i i2 = −1

i3 = −i i4 = 1 i5 = i i6 = −1 i7 = −i i8 = 1.


(1 + i)2 = 1 + 2i + i2 = 1 + 2i + (−1) = 2i

(1 + i)8 = ((1 + i)2)4 = 24i4 = 24 = 16

I Note that


2+

(ix)3

6+

(ix)4

24+

(ix)5

120+ · · ·

= 1 + ix −x2

2− i

x3

6+

x4

24+ i

x5

120+ · · ·

=

(1−

x2

2+

x4

24+ · · ·

)+

(x −

x3

6+

x5

120+ · · ·

)i






i0 = 1 i1 = i i2 = −1 i3 = −i

i4 = 1 i5 = i i6 = −1 i7 = −i i8 = 1.


(1 + i)2 = 1 + 2i + i2 = 1 + 2i + (−1) = 2i

(1 + i)8 = ((1 + i)2)4 = 24i4 = 24 = 16

I Note that


2+

(ix)3

6+

(ix)4

24+

(ix)5

120+ · · ·

= 1 + ix −x2

2− i

x3

6+

x4

24+ i

x5

120+ · · ·

=

(1−

x2

2+

x4

24+ · · ·

)+

(x −

x3

6+

x5

120+ · · ·

)i






i0 = 1 i1 = i i2 = −1 i3 = −i i4 = 1

i5 = i i6 = −1 i7 = −i i8 = 1.


(1 + i)2 = 1 + 2i + i2 = 1 + 2i + (−1) = 2i

(1 + i)8 = ((1 + i)2)4 = 24i4 = 24 = 16

I Note that


2+

(ix)3

6+

(ix)4

24+

(ix)5

120+ · · ·

= 1 + ix −x2

2− i

x3

6+

x4

24+ i

x5

120+ · · ·

=

(1−

x2

2+

x4

24+ · · ·

)+

(x −

x3

6+

x5

120+ · · ·

)i






i0 = 1 i1 = i i2 = −1 i3 = −i i4 = 1 i5 = i

i6 = −1 i7 = −i i8 = 1.


(1 + i)2 = 1 + 2i + i2 = 1 + 2i + (−1) = 2i

(1 + i)8 = ((1 + i)2)4 = 24i4 = 24 = 16

I Note that


2+

(ix)3

6+

(ix)4

24+

(ix)5

120+ · · ·

= 1 + ix −x2

2− i

x3

6+

x4

24+ i

x5

120+ · · ·

=

(1−

x2

2+

x4

24+ · · ·

)+

(x −

x3

6+

x5

120+ · · ·

)i






i0 = 1 i1 = i i2 = −1 i3 = −i i4 = 1 i5 = i i6 = −1

i7 = −i i8 = 1.


(1 + i)2 = 1 + 2i + i2 = 1 + 2i + (−1) = 2i

(1 + i)8 = ((1 + i)2)4 = 24i4 = 24 = 16

I Note that


2+

(ix)3

6+

(ix)4

24+

(ix)5

120+ · · ·

= 1 + ix −x2

2− i

x3

6+

x4

24+ i

x5

120+ · · ·

=

(1−

x2

2+

x4

24+ · · ·

)+

(x −

x3

6+

x5

120+ · · ·

)i






i0 = 1 i1 = i i2 = −1 i3 = −i i4 = 1 i5 = i i6 = −1 i7 = −i

i8 = 1.


(1 + i)2 = 1 + 2i + i2 = 1 + 2i + (−1) = 2i

(1 + i)8 = ((1 + i)2)4 = 24i4 = 24 = 16

I Note that


2+

(ix)3

6+

(ix)4

24+

(ix)5

120+ · · ·

= 1 + ix −x2

2− i

x3

6+

x4

24+ i

x5

120+ · · ·

=

(1−

x2

2+

x4

24+ · · ·

)+

(x −

x3

6+

x5

120+ · · ·

)i






i0 = 1 i1 = i i2 = −1 i3 = −i i4 = 1 i5 = i i6 = −1 i7 = −i i8 = 1.


(1 + i)2 = 1 + 2i + i2 = 1 + 2i + (−1) = 2i

(1 + i)8 = ((1 + i)2)4 = 24i4 = 24 = 16

I Note that


2+

(ix)3

6+

(ix)4

24+

(ix)5

120+ · · ·

= 1 + ix −x2

2− i

x3

6+

x4

24+ i

x5

120+ · · ·

=

(1−

x2

2+

x4

24+ · · ·

)+

(x −

x3

6+

x5

120+ · · ·

)i






i0 = 1 i1 = i i2 = −1 i3 = −i i4 = 1 i5 = i i6 = −1 i7 = −i i8 = 1.


(1 + i)2 = 1 + 2i + i2 = 1 + 2i + (−1) = 2i

(1 + i)8 = ((1 + i)2)4 = 24i4 = 24 = 16

I Note that


2+

(ix)3

6+

(ix)4

24+

(ix)5

120+ · · ·

= 1 + ix −x2

2− i

x3

6+

x4

24+ i

x5

120+ · · ·

=

(1−

x2

2+

x4

24+ · · ·

)+

(x −

x3

6+

x5

120+ · · ·

)i






i0 = 1 i1 = i i2 = −1 i3 = −i i4 = 1 i5 = i i6 = −1 i7 = −i i8 = 1.


(1 + i)2 = 1 + 2i + i2 = 1 + 2i + (−1) = 2i

(1 + i)8 = ((1 + i)2)4 = 24i4 = 24 = 16

I Note that


2+

(ix)3

6+

(ix)4

24+

(ix)5

120+ · · ·

= 1 + ix −x2

2− i

x3

6+

x4

24+ i

x5

120+ · · ·

=

(1−

x2

2+

x4

24+ · · ·

)+

(x −

x3

6+

x5

120+ · · ·

)i






i0 = 1 i1 = i i2 = −1 i3 = −i i4 = 1 i5 = i i6 = −1 i7 = −i i8 = 1.


(1 + i)2 = 1 + 2i + i2 = 1 + 2i + (−1) = 2i

(1 + i)8 = ((1 + i)2)4 = 24i4 = 24 = 16

I Note that


2+

(ix)3

6+

(ix)4

24+

(ix)5

120+ · · ·

= 1 + ix −x2

2− i

x3

6+

x4

24+ i

x5

120+ · · ·

=

(1−

x2

2+

x4

24+ · · ·

)+

(x −

x3

6+

x5

120+ · · ·

)i






i0 = 1 i1 = i i2 = −1 i3 = −i i4 = 1 i5 = i i6 = −1 i7 = −i i8 = 1.


(1 + i)2 = 1 + 2i + i2 = 1 + 2i + (−1) = 2i

(1 + i)8 = ((1 + i)2)4 = 24i4 = 24 = 16

I Note that


2+

(ix)3

6+

(ix)4

24+

(ix)5

120+ · · ·

= 1 + ix −x2

2− i

x3

6+

x4

24+ i

x5

120+ · · ·

=

(1−

x2

2+

x4

24+ · · ·

)+

(x −

x3

6+

x5

120+ · · ·

)i






i0 = 1 i1 = i i2 = −1 i3 = −i i4 = 1 i5 = i i6 = −1 i7 = −i i8 = 1.


(1 + i)2 = 1 + 2i + i2 = 1 + 2i + (−1) = 2i

(1 + i)8 = ((1 + i)2)4 = 24i4 = 24 = 16

I Note that


2+

(ix)3

6+

(ix)4

24+

(ix)5

120+ · · ·

= 1 + ix −x2

2− i

x3

6+

x4

24+ i

x5

120+ · · ·

=

(1−

x2

2+

x4

24+ · · ·

)+

(x −

x3

6+

x5

120+ · · ·

)i


De Moivre’s theorem

e iθ = exp(iθ) = cos(θ) + sin(θ)i

e−iθ = exp(−iθ) = cos(θ)− sin(θ)i

sin(θ)

cos(θ)

tan(θ) =sin(θ)

cos(θ)=

sinh(iθ)/i

cosh(iθ)= tanh(iθ)/i .

cos(a)2 + sin(a)2 = 1sec(a)2 = 1 + tan(a)2

sin(a + b) = sin(a) cos(b) + cos(a) sin(b)cos(a + b) = cos(a) cos(b)− sin(a) sin(b)

sin(2a) = 2 sin(a) cos(a)cos(2a) = 2 cos(a)2 − 1 = 1− 2 sin(a)2.




sin(θ)

cos(θ)

tan(θ) =sin(θ)

cos(θ)=

sinh(iθ)/i








sin(θ)

cos(θ)

tan(θ) =sin(θ)

cos(θ)=

sinh(iθ)/i








sin(θ) =e iθ − e−iθ

2i

cos(θ)

tan(θ) =sin(θ)

cos(θ)=

sinh(iθ)/i









2i= sinh(iθ)/i

cos(θ)

tan(θ) =sin(θ)

cos(θ)=

sinh(iθ)/i









2i= sinh(iθ)/i

cos(θ) =e iθ + e−iθ

2

tan(θ) =sin(θ)

cos(θ)=

sinh(iθ)/i









2i= sinh(iθ)/i


2= cosh(iθ)

tan(θ) =sin(θ)

cos(θ)=

sinh(iθ)/i









2i= sinh(iθ)/i


2= cosh(iθ)

tan(θ) =sin(θ)

cos(θ)

=sinh(iθ)/i









2i= sinh(iθ)/i


2= cosh(iθ)

tan(θ) =sin(θ)

cos(θ)=

sinh(iθ)/i

cosh(iθ)

= tanh(iθ)/i .








2i= sinh(iθ)/i


2= cosh(iθ)

tan(θ) =sin(θ)

cos(θ)=

sinh(iθ)/i









2i= sinh(iθ)/i


2= cosh(iθ)

tan(θ) =sin(θ)

cos(θ)=

sinh(iθ)/i


cos(a)2 + sin(a)2 = 1

sec(a)2 = 1 + tan(a)2







2i= sinh(iθ)/i


2= cosh(iθ)

tan(θ) =sin(θ)

cos(θ)=

sinh(iθ)/i









2i= sinh(iθ)/i


2= cosh(iθ)

tan(θ) =sin(θ)

cos(θ)=

sinh(iθ)/i



sin(a + b) = sin(a) cos(b) + cos(a) sin(b)

cos(a + b) = cos(a) cos(b)− sin(a) sin(b)sin(2a) = 2 sin(a) cos(a)cos(2a) = 2 cos(a)2 − 1 = 1− 2 sin(a)2.





2i= sinh(iθ)/i


2= cosh(iθ)

tan(θ) =sin(θ)

cos(θ)=

sinh(iθ)/i









2i= sinh(iθ)/i


2= cosh(iθ)

tan(θ) =sin(θ)

cos(θ)=

sinh(iθ)/i




sin(2a) = 2 sin(a) cos(a)

cos(2a) = 2 cos(a)2 − 1 = 1− 2 sin(a)2.





2i= sinh(iθ)/i


2= cosh(iθ)

tan(θ) =sin(θ)

cos(θ)=

sinh(iθ)/i





Examples

cos(a)2 + sin(a)2 =

(e ia + e−ia

2

)2

+

(e ia − e−ia

2i

)2

= 2/4− 2/(−4) = 1

cos(a)2 − sin(a)2 =

(e ia + e−ia

2

)2

−(e ia − e−ia

2i

)2

= (e2ia + 2 + e−2ia)/4 + (e2ia − 2 + e−2ia)/4

= cos(2a)

2 sin(a) cos(a) = 2

(e ia − e−ia

2i

)(e ia + e−ia

2

)=

2

4i

(e2ia + e0 − e0 − e−2ia

)= (e2ia − e−2ia)/(2i) = sin(2a)

Examples

cos(a)2 + sin(a)2

=

(e ia + e−ia

2

)2

+

(e ia − e−ia

2i

)2

= 2/4− 2/(−4) = 1


(e ia + e−ia

2

)2

−(e ia − e−ia

2i

)2

= (e2ia + 2 + e−2ia)/4 + (e2ia − 2 + e−2ia)/4

= cos(2a)

2 sin(a) cos(a) = 2

(e ia − e−ia

2i

)(e ia + e−ia

2

)=

2

4i

(e2ia + e0 − e0 − e−2ia

)= (e2ia − e−2ia)/(2i) = sin(2a)

Examples

cos(a)2 + sin(a)2 =

(e ia + e−ia

2

)2

+

(e ia − e−ia

2i

)2

= 2/4− 2/(−4) = 1


(e ia + e−ia

2

)2

−(e ia − e−ia

2i

)2

= (e2ia + 2 + e−2ia)/4 + (e2ia − 2 + e−2ia)/4

= cos(2a)

2 sin(a) cos(a) = 2

(e ia − e−ia

2i

)(e ia + e−ia

2

)=

2

4i

(e2ia + e0 − e0 − e−2ia

)= (e2ia − e−2ia)/(2i) = sin(2a)

Examples

cos(a)2 + sin(a)2 =

(e ia + e−ia

2

)2

+

(e ia − e−ia

2i

)2

= (e2ia + 2e ia−ia + e−2ia)/4 + (e2ia − 2e ia−ia + e−2ia)/(−4)

= 2/4− 2/(−4) = 1


(e ia + e−ia

2

)2

−(e ia − e−ia

2i

)2

= (e2ia + 2 + e−2ia)/4 + (e2ia − 2 + e−2ia)/4

= cos(2a)

2 sin(a) cos(a) = 2

(e ia − e−ia

2i

)(e ia + e−ia

2

)=

2

4i

(e2ia + e0 − e0 − e−2ia

)= (e2ia − e−2ia)/(2i) = sin(2a)

Examples

cos(a)2 + sin(a)2 =

(e ia + e−ia

2

)2

+

(e ia − e−ia

2i

)2

= (e2ia + 2 + e−2ia)/4 + (e2ia − 2 + e−2ia)/(−4)

= 2/4− 2/(−4) = 1


(e ia + e−ia

2

)2

−(e ia − e−ia

2i

)2

= (e2ia + 2 + e−2ia)/4 + (e2ia − 2 + e−2ia)/4

= cos(2a)

2 sin(a) cos(a) = 2

(e ia − e−ia

2i

)(e ia + e−ia

2

)=

2

4i

(e2ia + e0 − e0 − e−2ia

)= (e2ia − e−2ia)/(2i) = sin(2a)

Examples

cos(a)2 + sin(a)2 =

(e ia + e−ia

2

)2

+

(e ia − e−ia

2i

)2

= (e2ia + 2 + e−2ia)/4 + (e2ia − 2 + e−2ia)/(−4)

= 2/4− 2/(−4) = 1


(e ia + e−ia

2

)2

−(e ia − e−ia

2i

)2

= (e2ia + 2 + e−2ia)/4 + (e2ia − 2 + e−2ia)/4

= cos(2a)

2 sin(a) cos(a) = 2

(e ia − e−ia

2i

)(e ia + e−ia

2

)=

2

4i

(e2ia + e0 − e0 − e−2ia

)= (e2ia − e−2ia)/(2i) = sin(2a)

Examples

cos(a)2 + sin(a)2 =

(e ia + e−ia

2

)2

+

(e ia − e−ia

2i

)2

= (e2ia + 2 + e−2ia)/4 + (e2ia − 2 + e−2ia)/(−4)

= 2/4− 2/(−4) = 1

cos(a)2 − sin(a)2

=

(e ia + e−ia

2

)2

−(e ia − e−ia

2i

)2

= (e2ia + 2 + e−2ia)/4 + (e2ia − 2 + e−2ia)/4

= cos(2a)

2 sin(a) cos(a) = 2

(e ia − e−ia

2i

)(e ia + e−ia

2

)=

2

4i

(e2ia + e0 − e0 − e−2ia

)= (e2ia − e−2ia)/(2i) = sin(2a)

Examples

cos(a)2 + sin(a)2 =

(e ia + e−ia

2

)2

+

(e ia − e−ia

2i

)2

= (e2ia + 2 + e−2ia)/4 + (e2ia − 2 + e−2ia)/(−4)

= 2/4− 2/(−4) = 1


(e ia + e−ia

2

)2

−(e ia − e−ia

2i

)2

= (e2ia + 2 + e−2ia)/4 + (e2ia − 2 + e−2ia)/4

= cos(2a)

2 sin(a) cos(a) = 2

(e ia − e−ia

2i

)(e ia + e−ia

2

)=

2

4i

(e2ia + e0 − e0 − e−2ia

)= (e2ia − e−2ia)/(2i) = sin(2a)

Examples

cos(a)2 + sin(a)2 =

(e ia + e−ia

2

)2

+

(e ia − e−ia

2i

)2

= (e2ia + 2 + e−2ia)/4 + (e2ia − 2 + e−2ia)/(−4)

= 2/4− 2/(−4) = 1


(e ia + e−ia

2

)2

−(e ia − e−ia

2i

)2

= (e2ia + 2 + e−2ia)/4 + (e2ia − 2 + e−2ia)/4

= cos(2a)

2 sin(a) cos(a) = 2

(e ia − e−ia

2i

)(e ia + e−ia

2

)=

2

4i

(e2ia + e0 − e0 − e−2ia

)= (e2ia − e−2ia)/(2i) = sin(2a)

Examples

cos(a)2 + sin(a)2 =

(e ia + e−ia

2

)2

+

(e ia − e−ia

2i

)2

= (e2ia + 2 + e−2ia)/4 + (e2ia − 2 + e−2ia)/(−4)

= 2/4− 2/(−4) = 1


(e ia + e−ia

2

)2

−(e ia − e−ia

2i

)2

= (e2ia + 2 + e−2ia)/4 + (e2ia − 2 + e−2ia)/4

= 2(e2ia + e−2ia)/4

= cos(2a)

2 sin(a) cos(a) = 2

(e ia − e−ia

2i

)(e ia + e−ia

2

)=

2

4i

(e2ia + e0 − e0 − e−2ia

)= (e2ia − e−2ia)/(2i) = sin(2a)

Examples

cos(a)2 + sin(a)2 =

(e ia + e−ia

2

)2

+

(e ia − e−ia

2i

)2

= (e2ia + 2 + e−2ia)/4 + (e2ia − 2 + e−2ia)/(−4)

= 2/4− 2/(−4) = 1


(e ia + e−ia

2

)2

−(e ia − e−ia

2i

)2

= (e2ia + 2 + e−2ia)/4 + (e2ia − 2 + e−2ia)/4

= (e2ia + e−2ia)/2

= cos(2a)

2 sin(a) cos(a) = 2

(e ia − e−ia

2i

)(e ia + e−ia

2

)=

2

4i

(e2ia + e0 − e0 − e−2ia

)= (e2ia − e−2ia)/(2i) = sin(2a)

Examples

cos(a)2 + sin(a)2 =

(e ia + e−ia

2

)2

+

(e ia − e−ia

2i

)2

= (e2ia + 2 + e−2ia)/4 + (e2ia − 2 + e−2ia)/(−4)

= 2/4− 2/(−4) = 1


(e ia + e−ia

2

)2

−(e ia − e−ia

2i

)2

= (e2ia + 2 + e−2ia)/4 + (e2ia − 2 + e−2ia)/4

= (e2ia + e−2ia)/2 = cos(2a)

2 sin(a) cos(a) = 2

(e ia − e−ia

2i

)(e ia + e−ia

2

)=

2

4i

(e2ia + e0 − e0 − e−2ia

)= (e2ia − e−2ia)/(2i) = sin(2a)

Examples

cos(a)2 + sin(a)2 =

(e ia + e−ia

2

)2

+

(e ia − e−ia

2i

)2

= (e2ia + 2 + e−2ia)/4 + (e2ia − 2 + e−2ia)/(−4)

= 2/4− 2/(−4) = 1


(e ia + e−ia

2

)2

−(e ia − e−ia

2i

)2

= (e2ia + 2 + e−2ia)/4 + (e2ia − 2 + e−2ia)/4

= (e2ia + e−2ia)/2 = cos(2a)

2 sin(a) cos(a)

= 2

(e ia − e−ia

2i

)(e ia + e−ia

2

)=

2

4i

(e2ia + e0 − e0 − e−2ia

)= (e2ia − e−2ia)/(2i) = sin(2a)

Examples

cos(a)2 + sin(a)2 =

(e ia + e−ia

2

)2

+

(e ia − e−ia

2i

)2

= (e2ia + 2 + e−2ia)/4 + (e2ia − 2 + e−2ia)/(−4)

= 2/4− 2/(−4) = 1


(e ia + e−ia

2

)2

−(e ia − e−ia

2i

)2

= (e2ia + 2 + e−2ia)/4 + (e2ia − 2 + e−2ia)/4

= (e2ia + e−2ia)/2 = cos(2a)

2 sin(a) cos(a) = 2

(e ia − e−ia

2i

)(e ia + e−ia

2

)

=2

4i

(e2ia + e0 − e0 − e−2ia

)= (e2ia − e−2ia)/(2i) = sin(2a)

Examples

cos(a)2 + sin(a)2 =

(e ia + e−ia

2

)2

+

(e ia − e−ia

2i

)2

= (e2ia + 2 + e−2ia)/4 + (e2ia − 2 + e−2ia)/(−4)

= 2/4− 2/(−4) = 1


(e ia + e−ia

2

)2

−(e ia − e−ia

2i

)2

= (e2ia + 2 + e−2ia)/4 + (e2ia − 2 + e−2ia)/4

= (e2ia + e−2ia)/2 = cos(2a)

2 sin(a) cos(a) = 2

(e ia − e−ia

2i

)(e ia + e−ia

2

)=

2

4i

(e2ia + e0 − e0 − e−2ia

)

= (e2ia − e−2ia)/(2i) = sin(2a)

Examples

cos(a)2 + sin(a)2 =

(e ia + e−ia

2

)2

+

(e ia − e−ia

2i

)2

= (e2ia + 2 + e−2ia)/4 + (e2ia − 2 + e−2ia)/(−4)

= 2/4− 2/(−4) = 1


(e ia + e−ia

2

)2

−(e ia − e−ia

2i

)2

= (e2ia + 2 + e−2ia)/4 + (e2ia − 2 + e−2ia)/4

= (e2ia + e−2ia)/2 = cos(2a)

2 sin(a) cos(a) = 2

(e ia − e−ia

2i

)(e ia + e−ia

2

)=

2

4i

(e2ia + e0 − e0 − e−2ia

)= (e2ia − e−2ia)/(2i)

= sin(2a)

Examples

cos(a)2 + sin(a)2 =

(e ia + e−ia

2

)2

+

(e ia − e−ia

2i

)2

= (e2ia + 2 + e−2ia)/4 + (e2ia − 2 + e−2ia)/(−4)

= 2/4− 2/(−4) = 1


(e ia + e−ia

2

)2

−(e ia − e−ia

2i

)2

= (e2ia + 2 + e−2ia)/4 + (e2ia − 2 + e−2ia)/4

= (e2ia + e−2ia)/2 = cos(2a)

2 sin(a) cos(a) = 2

(e ia − e−ia

2i

)(e ia + e−ia

2

)=

2

4i

(e2ia + e0 − e0 − e−2ia

)= (e2ia − e−2ia)/(2i) = sin(2a)

The addition formula


0

0

0 0

sin(a) cos(b) + cos(a) sin(b) =e ia − e−ia

2i

e ib + e−ib

2+

e ia + e−ia

2

e ib − e−ib

2i

=e i(a+b) − e−i(a+b)

2i= sin(a + b)



0

0

0 0b

cos(b)

sin(b)


2i

e ib + e−ib

2+

e ia + e−ia

2

e ib − e−ib

2i


2i= sin(a + b)



0

0

0 0

b cos(b)

sin(b)

a


2i

e ib + e−ib

2+

e ia + e−ia

2

e ib − e−ib

2i


2i= sin(a + b)



0

0

0 0

b cos(b)

sin(b)

a

sin(a) cos(b)


2i

e ib + e−ib

2+

e ia + e−ia

2

e ib − e−ib

2i


2i= sin(a + b)



0

0

0 0

b cos(b)

sin(b)

a

sin(a) cos(b)


2i

e ib + e−ib

2+

e ia + e−ia

2

e ib − e−ib

2i


2i= sin(a + b)



0

0

0 0

b cos(b)

sin(b)

a

sin(a) cos(b)

a

π/2−a


2i

e ib + e−ib

2+

e ia + e−ia

2

e ib − e−ib

2i


2i= sin(a + b)



0

0

0 0

b cos(b)

sin(b)

a

sin(a) cos(b)

a

π/2−a

cos(a) sin(b)


2i

e ib + e−ib

2+

e ia + e−ia

2

e ib − e−ib

2i


2i= sin(a + b)



0

0

0 0

b cos(b)

sin(b)

a

sin(a) cos(b)

a

π/2−a

cos(a) sin(b)


2i

e ib + e−ib

2+

e ia + e−ia

2

e ib − e−ib

2i


2i= sin(a + b)



0

0

0 0

a+b

sin(a+b)

sin(a) cos(b)

cos(a) sin(b)


2i

e ib + e−ib

2+

e ia + e−ia

2

e ib − e−ib

2i


2i= sin(a + b)



0

0

0 0

a+b

sin(a+b)

sin(a) cos(b)

cos(a) sin(b)


2i

e ib + e−ib

2+

e ia + e−ia

2

e ib − e−ib

2i


2i= sin(a + b)



0

0

0 0

a+b

sin(a+b)

sin(a) cos(b)

cos(a) sin(b)


2i

e ib + e−ib

2+

e ia + e−ia

2

e ib − e−ib

2i


2i

= sin(a + b)



0

0

0 0

a+b

sin(a+b)

sin(a) cos(b)

cos(a) sin(b)


2i

e ib + e−ib

2+

e ia + e−ia

2

e ib − e−ib

2i


2i= sin(a + b)

Finite Fourier series

I A finite Fourier series is a sum of constant multiples of functions of theform sin(nx) or cos(mx) (with n,m ∈ Z). Note that the constant functionf (x) = a = a cos(0x) is included.

I The phrase trigonometric polynomial means the same thing.

I Many functions can be rewritten as finite Fourier series:

sin(x)2 = 12− 1

2cos(2x)

sin(x)3 = 34sin(x)− 1

4sin(3x)

sin(x) sin(2x) sin(4x) = −sin(x)/4 + sin(3x)/4 + sin(5x)/4− sin(7x)/4

sin(x)4 + cos(x)4 = 34

+ 14cos(4x)

sin(nx) sin(mx) = 12cos((n −m)x)− 1

2cos((n + m)x).

I Method: Rewrite using cos(nθ) = (e inθ + e−inθ)/2 andsin(nθ) = (e inθ − e−inθ)/2i , expand out, then rewrite usinge imθ = cos(mθ) + sin(mθ)i .

I Once a function has been rewritten in this form, it is very easy todifferentiate it or integrate it.





sin(x)2 = 12− 1

2cos(2x)

sin(x)3 = 34sin(x)− 1

4sin(3x)


sin(x)4 + cos(x)4 = 34

+ 14cos(4x)


2cos((n + m)x).







sin(x)2 = 12− 1

2cos(2x)

sin(x)3 = 34sin(x)− 1

4sin(3x)


sin(x)4 + cos(x)4 = 34

+ 14cos(4x)


2cos((n + m)x).







sin(x)2 = 12− 1

2cos(2x)

sin(x)3 = 34sin(x)− 1

4sin(3x)


sin(x)4 + cos(x)4 = 34

+ 14cos(4x)


2cos((n + m)x).







sin(x)2 = 12− 1

2cos(2x)

sin(x)3 = 34sin(x)− 1

4sin(3x)


sin(x)4 + cos(x)4 = 34

+ 14cos(4x)


2cos((n + m)x).







sin(x)2 = 12− 1

2cos(2x)

sin(x)3 = 34sin(x)− 1

4sin(3x)


sin(x)4 + cos(x)4 = 34

+ 14cos(4x)


2cos((n + m)x).







sin(x)2 = 12− 1

2cos(2x)

sin(x)3 = 34sin(x)− 1

4sin(3x)


sin(x)4 + cos(x)4 = 34

+ 14cos(4x)


2cos((n + m)x).







sin(x)2 = 12− 1

2cos(2x)

sin(x)3 = 34sin(x)− 1

4sin(3x)


sin(x)4 + cos(x)4 = 34

+ 14cos(4x)


2cos((n + m)x).







sin(x)2 = 12− 1

2cos(2x)

sin(x)3 = 34sin(x)− 1

4sin(3x)


sin(x)4 + cos(x)4 = 34

+ 14cos(4x)


2cos((n + m)x).







sin(x)2 = 12− 1

2cos(2x)

sin(x)3 = 34sin(x)− 1

4sin(3x)


sin(x)4 + cos(x)4 = 34

+ 14cos(4x)


2cos((n + m)x).







sin(x)2 = 12− 1

2cos(2x)

sin(x)3 = 34sin(x)− 1

4sin(3x)


sin(x)4 + cos(x)4 = 34

+ 14cos(4x)


2cos((n + m)x).



Examples

Problem: write sin(x)4 + cos(x)4 as a Fourier series.

Put u = e ix , so sin(x) = (u − u−1)/(2i) and cos(x) = (u + u−1)/2. Note thati2 = −1 so i4 = (−1)2 = 1 so (2i)4 = 24 = 16. Note also that

(x + y)4 = x4 + 4x3y + 6x2y 2 + 4xy 3 + y 4

(use the binomial formula, or expand it out.) Thus

sin(x)4 + cos(x)4 = (u − u−1)4/16 + (u + u−1)4/16

= (u4 − 4u2 + 6− 4u−2 + u−4)/16 +

(u4 + 4u2 + 6 + 4u−2 + u−4)/16

= 12/16 + 2(u4 + u−4)/16 = 3/4 + ((u4 + u−4)/2)/4

= (3 + cos(4x))/4

Examples



(x + y)4 = x4 + 4x3y + 6x2y 2 + 4xy 3 + y 4


sin(x)4 + cos(x)4 = (u − u−1)4/16 + (u + u−1)4/16

= (u4 − 4u2 + 6− 4u−2 + u−4)/16 +

(u4 + 4u2 + 6 + 4u−2 + u−4)/16

= 12/16 + 2(u4 + u−4)/16 = 3/4 + ((u4 + u−4)/2)/4

= (3 + cos(4x))/4

Examples


Put u = e ix , so sin(x) = (u − u−1)/(2i) and cos(x) = (u + u−1)/2.

Note thati2 = −1 so i4 = (−1)2 = 1 so (2i)4 = 24 = 16. Note also that

(x + y)4 = x4 + 4x3y + 6x2y 2 + 4xy 3 + y 4


sin(x)4 + cos(x)4 = (u − u−1)4/16 + (u + u−1)4/16

= (u4 − 4u2 + 6− 4u−2 + u−4)/16 +

(u4 + 4u2 + 6 + 4u−2 + u−4)/16

= 12/16 + 2(u4 + u−4)/16 = 3/4 + ((u4 + u−4)/2)/4

= (3 + cos(4x))/4

Examples


Put u = e ix , so sin(x) = (u − u−1)/(2i) and cos(x) = (u + u−1)/2. Note thati2 = −1 so i4 = (−1)2 = 1 so (2i)4 = 24 = 16.

Note also that

(x + y)4 = x4 + 4x3y + 6x2y 2 + 4xy 3 + y 4


sin(x)4 + cos(x)4 = (u − u−1)4/16 + (u + u−1)4/16

= (u4 − 4u2 + 6− 4u−2 + u−4)/16 +

(u4 + 4u2 + 6 + 4u−2 + u−4)/16

= 12/16 + 2(u4 + u−4)/16 = 3/4 + ((u4 + u−4)/2)/4

= (3 + cos(4x))/4

Examples



(x + y)4 = x4 + 4x3y + 6x2y 2 + 4xy 3 + y 4


sin(x)4 + cos(x)4 = (u − u−1)4/16 + (u + u−1)4/16

= (u4 − 4u2 + 6− 4u−2 + u−4)/16 +

(u4 + 4u2 + 6 + 4u−2 + u−4)/16

= 12/16 + 2(u4 + u−4)/16 = 3/4 + ((u4 + u−4)/2)/4

= (3 + cos(4x))/4

Examples



(x + y)4 = x4 + 4x3y + 6x2y 2 + 4xy 3 + y 4

(use the binomial formula, or expand it out.)

Thus

sin(x)4 + cos(x)4 = (u − u−1)4/16 + (u + u−1)4/16

= (u4 − 4u2 + 6− 4u−2 + u−4)/16 +

(u4 + 4u2 + 6 + 4u−2 + u−4)/16

= 12/16 + 2(u4 + u−4)/16 = 3/4 + ((u4 + u−4)/2)/4

= (3 + cos(4x))/4

Examples



(x + y)4 = x4 + 4x3y + 6x2y 2 + 4xy 3 + y 4


sin(x)4 + cos(x)4 = (u − u−1)4/16 + (u + u−1)4/16

= (u4 − 4u2 + 6− 4u−2 + u−4)/16 +

(u4 + 4u2 + 6 + 4u−2 + u−4)/16

= 12/16 + 2(u4 + u−4)/16 = 3/4 + ((u4 + u−4)/2)/4

= (3 + cos(4x))/4

Examples



(x + y)4 = x4 + 4x3y + 6x2y 2 + 4xy 3 + y 4


sin(x)4 + cos(x)4 = (u − u−1)4/16 + (u + u−1)4/16

= (u4 − 4u2 + 6− 4u−2 + u−4)/16 +

(u4 + 4u2 + 6 + 4u−2 + u−4)/16

= 12/16 + 2(u4 + u−4)/16 = 3/4 + ((u4 + u−4)/2)/4

= (3 + cos(4x))/4

Examples



(x + y)4 = x4 + 4x3y + 6x2y 2 + 4xy 3 + y 4


sin(x)4 + cos(x)4 = (u − u−1)4/16 + (u + u−1)4/16

= (u4 − 4u2 + 6− 4u−2 + u−4)/16 +

(u4 + 4u2 + 6 + 4u−2 + u−4)/16

= 12/16 + 2(u4 + u−4)/16

= 3/4 + ((u4 + u−4)/2)/4

= (3 + cos(4x))/4

Examples



(x + y)4 = x4 + 4x3y + 6x2y 2 + 4xy 3 + y 4


sin(x)4 + cos(x)4 = (u − u−1)4/16 + (u + u−1)4/16

= (u4 − 4u2 + 6− 4u−2 + u−4)/16 +

(u4 + 4u2 + 6 + 4u−2 + u−4)/16

= 12/16 + 2(u4 + u−4)/16 = 3/4 + ((u4 + u−4)/2)/4

= (3 + cos(4x))/4

Examples



(x + y)4 = x4 + 4x3y + 6x2y 2 + 4xy 3 + y 4


sin(x)4 + cos(x)4 = (u − u−1)4/16 + (u + u−1)4/16

= (u4 − 4u2 + 6− 4u−2 + u−4)/16 +

(u4 + 4u2 + 6 + 4u−2 + u−4)/16

= 12/16 + 2(u4 + u−4)/16 = 3/4 + ((u4 + u−4)/2)/4

= (3 + cos(4x))/4

Special values

You should know the following values of sin(θ) and cos(θ):

θ sin(θ) cos(θ) tan(θ)π/2 1 0 ∞π/3

√3/2 1/2

√3

π/4√

2/2√

2/2 1

π/6 1/2√

3/2√

3/3

Proved by considering these triangles:

1

1

√2

π/4

π/4π/2

1 1

12

12

√3

2π3

π3

π6π6

You should also be able to deduce things like cos(5π/6) = −√

3/2.

Special values



√3/2 1/2

√3

π/4√

2/2√

2/2 1

π/6 1/2√

3/2√

3/3


1

1

√2

π/4

π/4π/2

1 1

12

12

√3

2π3

π3

π6π6


3/2.

Special values



√3/2 1/2

√3

π/4√

2/2√

2/2 1

π/6 1/2√

3/2√

3/3


1

1

√2

π/4

π/4π/2

1 1

12

12

√3

2π3

π3

π6π6


3/2.

Special values



√3/2 1/2

√3

π/4√

2/2√

2/2 1

π/6 1/2√

3/2√

3/3


1

1

√2

π/4

π/4π/2

1 1

12

12

√3

2π3

π3

π6π6


3/2.

Inverse trigonometric functions

−1

1

−π2

π2

−1

1

π

−π2

π2

sin : [−π2,π

2]−→[−1,1] cos : [0,π]−→[−1,1] tan : [−π

2,π

2]−→R

−1 1

−π2

π2

−1 1

π

−π2

π2

arcsin : [−1,1]−→[−π2,π

2] arccos : [−1,1]−→[0,π] arctan : R−→[−π

2,π

2]

Differentiation


I The meaning of differentiation (slopes of graphs, time-dependent andspace-dependent variables, etc)

I Some derivatives from first principles: x2, 1/x , ex .I Rules for finding derivatives:

I The product rule ((uv)′ = u′v + uv ′)I The quotient rule ((u/v)′ = (u′v − uv ′)/v2)I The chain rule ( dz

dx= dz

dydydx

)

I The power rule ((un)′ = nun−1u′)I The logarithmic rule (log(u)′ = u′/u)I The inverse function rule ( dx

dy= 1/ dy

dx)

I Derivatives of various classes of functions (eg the derivative of a rationalfunction is another rational function.)

You must learn to find derivatives quickly and accurately.

Differentiation





dx= dz

dydydx

)


dy= 1/ dy

dx)



Differentiation





dx= dz

dydydx

)


dy= 1/ dy

dx)



Differentiation



I Some derivatives from first principles: x2, 1/x , ex .

I Rules for finding derivatives:


dx= dz

dydydx

)


dy= 1/ dy

dx)



Differentiation





dx= dz

dydydx

)


dy= 1/ dy

dx)



Differentiation




I The product rule ((uv)′ = u′v + uv ′)

I The quotient rule ((u/v)′ = (u′v − uv ′)/v2)I The chain rule ( dz

dx= dz

dydydx

)


dy= 1/ dy

dx)



Differentiation




I The product rule ((uv)′ = u′v + uv ′)I The quotient rule ((u/v)′ = (u′v − uv ′)/v2)

I The chain rule ( dzdx

= dzdy

dydx

)


dy= 1/ dy

dx)



Differentiation





dx= dz

dydydx

)


dy= 1/ dy

dx)



Differentiation





dx= dz

dydydx

)

I The power rule ((un)′ = nun−1u′)

I The logarithmic rule (log(u)′ = u′/u)I The inverse function rule ( dx

dy= 1/ dy

dx)



Differentiation





dx= dz

dydydx

)

I The power rule ((un)′ = nun−1u′)I The logarithmic rule (log(u)′ = u′/u)

I The inverse function rule ( dxdy

= 1/ dydx

)



Differentiation





dx= dz

dydydx

)


dy= 1/ dy

dx)



Differentiation





dx= dz

dydydx

)


dy= 1/ dy

dx)



Differentiation





dx= dz

dydydx

)


dy= 1/ dy

dx)



Meaning

I Consider related variables x and y ; so whenever x changes, so does y .I Examples:

I p = price of chocolate ; d = demand for chocolate .I t = time ; d = atmospheric CO2 concentration .I r = distance from sun ; g = strength of solar gravity .

I If x changes to x + δx , then y changes to y + δy .

dy

dx= limδx−→0

δy

δx= derivative of y with respect to x .

I If y = f (x), then δy = f (x + δx)− f (x), so

f ′(x) =dy

dx= limδx−→0

f (x + δx)− f (x)

δx

= limh−→0

f (x + h)− f (x)

h.

I We sometimes write y ′ for dy/dx (care needed).

Meaning

I Consider related variables x and y ; so whenever x changes, so does y .

I Examples:



dy

dx= limδx−→0

δy



f ′(x) =dy

dx= limδx−→0

f (x + δx)− f (x)

δx

= limh−→0

f (x + h)− f (x)

h.


Meaning




dy

dx= limδx−→0

δy



f ′(x) =dy

dx= limδx−→0

f (x + δx)− f (x)

δx

= limh−→0

f (x + h)− f (x)

h.


Meaning


I p = price of chocolate ; d = demand for chocolate .

I t = time ; d = atmospheric CO2 concentration .I r = distance from sun ; g = strength of solar gravity .


dy

dx= limδx−→0

δy



f ′(x) =dy

dx= limδx−→0

f (x + δx)− f (x)

δx

= limh−→0

f (x + h)− f (x)

h.


Meaning


I p = price of chocolate ; d = demand for chocolate .I t = time ; d = atmospheric CO2 concentration .

I r = distance from sun ; g = strength of solar gravity .


dy

dx= limδx−→0

δy



f ′(x) =dy

dx= limδx−→0

f (x + δx)− f (x)

δx

= limh−→0

f (x + h)− f (x)

h.


Meaning




dy

dx= limδx−→0

δy



f ′(x) =dy

dx= limδx−→0

f (x + δx)− f (x)

δx

= limh−→0

f (x + h)− f (x)

h.


Meaning




dy

dx= limδx−→0

δy



f ′(x) =dy

dx= limδx−→0

f (x + δx)− f (x)

δx

= limh−→0

f (x + h)− f (x)

h.


Meaning




dy

dx= limδx−→0

δy



f ′(x) =dy

dx= limδx−→0

f (x + δx)− f (x)

δx

= limh−→0

f (x + h)− f (x)

h.


Meaning




dy

dx= limδx−→0

δy



f ′(x) =dy

dx= limδx−→0

f (x + δx)− f (x)

δx

= limh−→0

f (x + h)− f (x)

h.


Meaning




dy

dx= limδx−→0

δy



f ′(x) =dy

dx= limδx−→0

f (x + δx)− f (x)

δx

= limh−→0

f (x + h)− f (x)

h.


Meaning




dy

dx= limδx−→0

δy



f ′(x) =dy

dx= limδx−→0

f (x + δx)− f (x)

δx= lim

h−→0

f (x + h)− f (x)

h.


Meaning




dy

dx= limδx−→0

δy



f ′(x) =dy

dx= limδx−→0

f (x + δx)− f (x)

δx= lim

h−→0

f (x + h)− f (x)

h.


Slopes

y=f (x)

0

Slopes

y=f (x)

x

y

Consider variables x and y related by y = f (x).

Slopes

y=f (x)

x

y

slope dy/dx

dy/dx is the slope of the tangent line to the graph.

Slopes

y=f (x)

x

y

slope dy/dx

x+δx

y+δy

If x changes by a small amount δx , then y will change by a small amountδy .

Slopes

y=f (x)

x

y

slope dy/dx

x+δx

y+δy

δx

δy

If x changes by a small amount δx , then y will change by a small amountδy .

Slopes

y=f (x)

x

y

slope dy/dx

x+δx

y+δy

δx

δy

slope δy/δx

The ratio δy/δx is the slope of a chord cutting across the graph.

Slopes

y=f (x)

x

y

slope dy/dx

x+δx

y+δy

δx

δy

slope δy/δx

The slope of the chord changes slightly as δx decreases.

Slopes

y=f (x)

x

y

slope dy/dx

x+δx

y+δy

δx

δy

slope δy/δx

As δx approaches zero, the chord approaches the tangent, and δy/δxapproaches dy/dx .

Slopes

y=f (x)

x

y

slope dy/dx

x+δx

y+δy

δx

δy

slope δy/δx


Slopes

y=f (x)

x

y

slope dy/dx


The function f (x) = x2

I Consider the function f (x) = x2.

I Then f (x + h) = (x + h)2 = x2 + 2xh + h2

, so

f (x + h)− f (x)

h=

(x + h)2 − x2

h

=x2 + 2xh + h2 − x2

h

=2xh + h2

h

= 2x + h

I Thus

f ′(x) = limh−→0

f (x + h)− f (x)

h

= limh−→0

(2x + h) = 2x .

I Similarly:ddx

(xn) = nxn−1 for all n.



I Then f (x + h) = (x + h)2 = x2 + 2xh + h2

, so

f (x + h)− f (x)

h=

(x + h)2 − x2

h

=x2 + 2xh + h2 − x2

h

=2xh + h2

h

= 2x + h

I Thus

f ′(x) = limh−→0

f (x + h)− f (x)

h

= limh−→0

(2x + h) = 2x .

I Similarly:ddx




I Then f (x + h) = (x + h)2 = x2 + 2xh + h2

, so

f (x + h)− f (x)

h=

(x + h)2 − x2

h

=x2 + 2xh + h2 − x2

h

=2xh + h2

h

= 2x + h

I Thus

f ′(x) = limh−→0

f (x + h)− f (x)

h

= limh−→0

(2x + h) = 2x .

I Similarly:ddx




I Then f (x + h) = (x + h)2 = x2 + 2xh + h2 , so

f (x + h)− f (x)

h=

(x + h)2 − x2

h

=x2 + 2xh + h2 − x2

h

=2xh + h2

h

= 2x + h

I Thus

f ′(x) = limh−→0

f (x + h)− f (x)

h

= limh−→0

(2x + h) = 2x .

I Similarly:ddx





f (x + h)− f (x)

h=

(x + h)2 − x2

h

=x2 + 2xh + h2 − x2

h

=2xh + h2

h

= 2x + h

I Thus

f ′(x) = limh−→0

f (x + h)− f (x)

h

= limh−→0

(2x + h) = 2x .

I Similarly:ddx





f (x + h)− f (x)

h=

(x + h)2 − x2

h

=x2 + 2xh + h2 − x2

h

=2xh + h2

h

= 2x + h

I Thus

f ′(x) = limh−→0

f (x + h)− f (x)

h

= limh−→0

(2x + h) = 2x .

I Similarly:ddx





f (x + h)− f (x)

h=

(x + h)2 − x2

h

=x2 + 2xh + h2 − x2

h

=2xh + h2

h

= 2x + h

I Thus

f ′(x) = limh−→0

f (x + h)− f (x)

h

= limh−→0

(2x + h) = 2x .

I Similarly:ddx





f (x + h)− f (x)

h=

(x + h)2 − x2

h

=x2 + 2xh + h2 − x2

h

=2xh + h2

h

= 2x + h

I Thus

f ′(x) = limh−→0

f (x + h)− f (x)

h

= limh−→0

(2x + h) = 2x .

I Similarly:ddx





f (x + h)− f (x)

h=

(x + h)2 − x2

h

=x2 + 2xh + h2 − x2

h

=2xh + h2

h

= 2x + h

I Thus

f ′(x) = limh−→0

f (x + h)− f (x)

h= lim

h−→0(2x + h)

= 2x .

I Similarly:ddx





f (x + h)− f (x)

h=

(x + h)2 − x2

h

=x2 + 2xh + h2 − x2

h

=2xh + h2

h

= 2x + h

I Thus

f ′(x) = limh−→0

f (x + h)− f (x)

h= lim

h−→0(2x + h) = 2x .

I Similarly:ddx





f (x + h)− f (x)

h=

(x + h)2 − x2

h

=x2 + 2xh + h2 − x2

h

=2xh + h2

h

= 2x + h

I Thus

f ′(x) = limh−→0

f (x + h)− f (x)

h= lim

h−→0(2x + h) = 2x .

I Similarly:ddx


The function f (x) = 1/x

I Consider the function f (x) = 1/x .

I . f (x + h)− f (x) =1

x + h− 1

x

=x − (x + h)

x(x + h)=

−hx(x + h)

sof (x + h)− f (x)

h=

−1

x(x + h)

so

f ′(x) = limh−→0

f (x + h)− f (x)

h

= limh−→0

−1

x(x + h)=−1

x2



I . f (x + h)− f (x) =1

x + h− 1

x

=x − (x + h)

x(x + h)=

−hx(x + h)

sof (x + h)− f (x)

h=

−1

x(x + h)

so

f ′(x) = limh−→0

f (x + h)− f (x)

h

= limh−→0

−1

x(x + h)=−1

x2



I . f (x + h)− f (x) =1

x + h− 1

x

=x − (x + h)

x(x + h)=

−hx(x + h)

sof (x + h)− f (x)

h=

−1

x(x + h)

so

f ′(x) = limh−→0

f (x + h)− f (x)

h

= limh−→0

−1

x(x + h)=−1

x2



I . f (x + h)− f (x) =1

x + h− 1

x=

x − (x + h)

x(x + h)

=−h

x(x + h)

sof (x + h)− f (x)

h=

−1

x(x + h)

so

f ′(x) = limh−→0

f (x + h)− f (x)

h

= limh−→0

−1

x(x + h)=−1

x2



I . f (x + h)− f (x) =1

x + h− 1

x=

x − (x + h)

x(x + h)=

−hx(x + h)

sof (x + h)− f (x)

h=

−1

x(x + h)

so

f ′(x) = limh−→0

f (x + h)− f (x)

h

= limh−→0

−1

x(x + h)=−1

x2



I . f (x + h)− f (x) =1

x + h− 1

x=

x − (x + h)

x(x + h)=

−hx(x + h)

sof (x + h)− f (x)

h=

−1

x(x + h)

so

f ′(x) = limh−→0

f (x + h)− f (x)

h

= limh−→0

−1

x(x + h)=−1

x2



I . f (x + h)− f (x) =1

x + h− 1

x=

x − (x + h)

x(x + h)=

−hx(x + h)

sof (x + h)− f (x)

h=

−1

x(x + h)

so

f ′(x) = limh−→0

f (x + h)− f (x)

h

= limh−→0

−1

x(x + h)=−1

x2



I . f (x + h)− f (x) =1

x + h− 1

x=

x − (x + h)

x(x + h)=

−hx(x + h)

sof (x + h)− f (x)

h=

−1

x(x + h)

so

f ′(x) = limh−→0

f (x + h)− f (x)

h= lim

h−→0

−1

x(x + h)

=−1

x2



I . f (x + h)− f (x) =1

x + h− 1

x=

x − (x + h)

x(x + h)=

−hx(x + h)

sof (x + h)− f (x)

h=

−1

x(x + h)

so

f ′(x) = limh−→0

f (x + h)− f (x)

h= lim

h−→0

−1

x(x + h)=−1

x2


I Consider the function f (x) = ex = 1 + x + x2

2!+ x3

3!+ · · · .

I .

f (x + h)− f (x) = ex+h − ex

= ex(eh − 1) = ex(h + h2

2!+ h3

3!+ · · ·

)so

f (x + h)− f (x)

h= ex

(1 + h

2!+ h2

3!+ · · ·

)so

f ′(x) = limh−→0

f (x + h)− f (x)

h

= limh−→0

ex(

1 + h2!

+ h2

3!+ · · ·

)= ex(1 + 0 + 0 + · · · )= ex .

I Conclusion: exp′(x) = exp(x).



2!+ x3

3!+ · · · .

I .

f (x + h)− f (x) = ex+h − ex

= ex(eh − 1) = ex(h + h2

2!+ h3

3!+ · · ·

)so

f (x + h)− f (x)

h= ex

(1 + h

2!+ h2

3!+ · · ·

)so

f ′(x) = limh−→0

f (x + h)− f (x)

h

= limh−→0

ex(

1 + h2!

+ h2

3!+ · · ·

)= ex(1 + 0 + 0 + · · · )= ex .




2!+ x3

3!+ · · · .

I .

f (x + h)− f (x) = ex+h − ex

= ex(eh − 1) = ex(h + h2

2!+ h3

3!+ · · ·

)so

f (x + h)− f (x)

h= ex

(1 + h

2!+ h2

3!+ · · ·

)so

f ′(x) = limh−→0

f (x + h)− f (x)

h

= limh−→0

ex(

1 + h2!

+ h2

3!+ · · ·

)= ex(1 + 0 + 0 + · · · )= ex .




2!+ x3

3!+ · · · .

I .

f (x + h)− f (x) = ex+h − ex = ex(eh − 1)

= ex(h + h2

2!+ h3

3!+ · · ·

)so

f (x + h)− f (x)

h= ex

(1 + h

2!+ h2

3!+ · · ·

)so

f ′(x) = limh−→0

f (x + h)− f (x)

h

= limh−→0

ex(

1 + h2!

+ h2

3!+ · · ·

)= ex(1 + 0 + 0 + · · · )= ex .




2!+ x3

3!+ · · · .

I .

f (x + h)− f (x) = ex+h − ex = ex(eh − 1) = ex(h + h2

2!+ h3

3!+ · · ·

)

sof (x + h)− f (x)

h= ex

(1 + h

2!+ h2

3!+ · · ·

)so

f ′(x) = limh−→0

f (x + h)− f (x)

h

= limh−→0

ex(

1 + h2!

+ h2

3!+ · · ·

)= ex(1 + 0 + 0 + · · · )= ex .




2!+ x3

3!+ · · · .

I .


2!+ h3

3!+ · · ·

)so

f (x + h)− f (x)

h= ex

(1 + h

2!+ h2

3!+ · · ·

)

so

f ′(x) = limh−→0

f (x + h)− f (x)

h

= limh−→0

ex(

1 + h2!

+ h2

3!+ · · ·

)= ex(1 + 0 + 0 + · · · )= ex .




2!+ x3

3!+ · · · .

I .


2!+ h3

3!+ · · ·

)so

f (x + h)− f (x)

h= ex

(1 + h

2!+ h2

3!+ · · ·

)so

f ′(x) = limh−→0

f (x + h)− f (x)

h

= limh−→0

ex(

1 + h2!

+ h2

3!+ · · ·

)= ex(1 + 0 + 0 + · · · )= ex .




2!+ x3

3!+ · · · .

I .


2!+ h3

3!+ · · ·

)so

f (x + h)− f (x)

h= ex

(1 + h

2!+ h2

3!+ · · ·

)so

f ′(x) = limh−→0

f (x + h)− f (x)

h

= limh−→0

ex(

1 + h2!

+ h2

3!+ · · ·

)

= ex(1 + 0 + 0 + · · · )= ex .




2!+ x3

3!+ · · · .

I .


2!+ h3

3!+ · · ·

)so

f (x + h)− f (x)

h= ex

(1 + h

2!+ h2

3!+ · · ·

)so

f ′(x) = limh−→0

f (x + h)− f (x)

h

= limh−→0

ex(

1 + h2!

+ h2

3!+ · · ·

)= ex(1 + 0 + 0 + · · · )

= ex .




2!+ x3

3!+ · · · .

I .


2!+ h3

3!+ · · ·

)so

f (x + h)− f (x)

h= ex

(1 + h

2!+ h2

3!+ · · ·

)so

f ′(x) = limh−→0

f (x + h)− f (x)

h

= limh−→0

ex(

1 + h2!

+ h2

3!+ · · ·

)= ex(1 + 0 + 0 + · · · )= ex .




2!+ x3

3!+ · · · .

I .


2!+ h3

3!+ · · ·

)so

f (x + h)− f (x)

h= ex

(1 + h

2!+ h2

3!+ · · ·

)so

f ′(x) = limh−→0

f (x + h)− f (x)

h

= limh−→0

ex(

1 + h2!

+ h2

3!+ · · ·

)= ex(1 + 0 + 0 + · · · )= ex .


Special functions

exp′(x) = exp(x) log′(x) = 1/x

sinh′(x) = cosh(x) arcsinh′(x) = (1 + x2)−1/2

cosh′(x) = sinh(x) arccosh′(x) = (x2 − 1)−1/2

tanh′(x) = sech(x)2 = 1− tanh(x)2 arctanh′(x) = (1− x2)−1

sin′(x) = cos(x) arcsin′(x) = (1− x2)−1/2

cos′(x) = − sin(x) arccos′(x) = −(1− x2)−1/2

tan′(x) = sec(x)2 = 1 + tan(x)2 arctan′(x) = (1 + x2)−1

I We showed earlier that exp′(x) = exp(x)

I We deduce sinh′(x) using the identity sinh(x) = (ex − e−x)/2. Similarlyfor cosh and tanh.

I Using cos(x) = cosh(ix) etc, we find sin′(x), cos′(x) and tan′(x).

I Using exp′(x) = exp(x) and the inverse function rule, we find thatlog′(x) = 1/x

I The inverse function rule also gives the remaining derivatives.

Special functions

exp′(x) = exp(x)

log′(x) = 1/x












Special functions

exp′(x) = exp(x)

log′(x) = 1/x

sinh′(x) = cosh(x)

arcsinh′(x) = (1 + x2)−1/2

cosh′(x) = sinh(x)

arccosh′(x) = (x2 − 1)−1/2

tanh′(x) = sech(x)2 = 1− tanh(x)2

arctanh′(x) = (1− x2)−1









Special functions

exp′(x) = exp(x)

log′(x) = 1/x


arcsinh′(x) = (1 + x2)−1/2


arccosh′(x) = (x2 − 1)−1/2


arctanh′(x) = (1− x2)−1

sin′(x) = cos(x)

arcsin′(x) = (1− x2)−1/2

cos′(x) = − sin(x)

arccos′(x) = −(1− x2)−1/2

tan′(x) = sec(x)2 = 1 + tan(x)2

arctan′(x) = (1 + x2)−1






Special functions



arcsinh′(x) = (1 + x2)−1/2


arccosh′(x) = (x2 − 1)−1/2


arctanh′(x) = (1− x2)−1

sin′(x) = cos(x)

arcsin′(x) = (1− x2)−1/2

cos′(x) = − sin(x)

arccos′(x) = −(1− x2)−1/2

tan′(x) = sec(x)2 = 1 + tan(x)2

arctan′(x) = (1 + x2)−1






Special functions













The product rule

I Consider variables u and v depending on x , and put w = uv . Then

w ′ = (uv)′ = u′v + uv ′

dwdx

= ddx

(uv) = dudxv + u dv

dx.

I If x changes to x + δx , then u changes to u + δu & v changes to v + δv

so w changes to

w + δw = (u + δu)(v + δv)

= uv + (δu) v + u (δv) + (δu)(δv)

δw = (δu) v + u (δv) + (δu)(δv)

δw

δx=δu

δxv + u

δv

δx+δu

δx

δv

δxδx

' du

dxv + u

dv

dx+

du

dx

dv

dxδx ' du

dxv + u

dv

dx

(The approximations become exact in the limit as δx −→ 0.)

The product rule


w ′ = (uv)′ = u′v + uv ′

dwdx

= ddx

(uv) = dudxv + u dv

dx.


so w changes to

w + δw = (u + δu)(v + δv)

= uv + (δu) v + u (δv) + (δu)(δv)

δw = (δu) v + u (δv) + (δu)(δv)

δw

δx=δu

δxv + u

δv

δx+δu

δx

δv

δxδx

' du

dxv + u

dv

dx+

du

dx

dv

dxδx ' du

dxv + u

dv

dx


The product rule


w ′ = (uv)′ = u′v + uv ′ dwdx

= ddx

(uv) = dudxv + u dv

dx.


so w changes to

w + δw = (u + δu)(v + δv)

= uv + (δu) v + u (δv) + (δu)(δv)

δw = (δu) v + u (δv) + (δu)(δv)

δw

δx=δu

δxv + u

δv

δx+δu

δx

δv

δxδx

' du

dxv + u

dv

dx+

du

dx

dv

dxδx ' du

dxv + u

dv

dx


The product rule


w ′ = (uv)′ = u′v + uv ′ dwdx

= ddx

(uv) = dudxv + u dv

dx.


so w changes to

w + δw = (u + δu)(v + δv)

= uv + (δu) v + u (δv) + (δu)(δv)

δw = (δu) v + u (δv) + (δu)(δv)

δw

δx=δu

δxv + u

δv

δx+δu

δx

δv

δxδx

' du

dxv + u

dv

dx+

du

dx

dv

dxδx ' du

dxv + u

dv

dx


The product rule


w ′ = (uv)′ = u′v + uv ′ dwdx

= ddx

(uv) = dudxv + u dv

dx.

I If x changes to x + δx , then u changes to u + δu & v changes to v + δvso w changes to

w + δw = (u + δu)(v + δv)

= uv + (δu) v + u (δv) + (δu)(δv)

δw = (δu) v + u (δv) + (δu)(δv)

δw

δx=δu

δxv + u

δv

δx+δu

δx

δv

δxδx

' du

dxv + u

dv

dx+

du

dx

dv

dxδx ' du

dxv + u

dv

dx


The product rule


w ′ = (uv)′ = u′v + uv ′ dwdx

= ddx

(uv) = dudxv + u dv

dx.


w + δw = (u + δu)(v + δv) = uv + (δu) v + u (δv) + (δu)(δv)

δw = (δu) v + u (δv) + (δu)(δv)

δw

δx=δu

δxv + u

δv

δx+δu

δx

δv

δxδx

' du

dxv + u

dv

dx+

du

dx

dv

dxδx ' du

dxv + u

dv

dx


The product rule


w ′ = (uv)′ = u′v + uv ′ dwdx

= ddx

(uv) = dudxv + u dv

dx.



δw = (δu) v + u (δv) + (δu)(δv)

δw

δx=δu

δxv + u

δv

δx+δu

δx

δv

δxδx

' du

dxv + u

dv

dx+

du

dx

dv

dxδx ' du

dxv + u

dv

dx


The product rule


w ′ = (uv)′ = u′v + uv ′ dwdx

= ddx

(uv) = dudxv + u dv

dx.



δw = (δu) v + u (δv) + (δu)(δv)

δw

δx=δu

δxv + u

δv

δx+δu

δx

δv

δxδx

' du

dxv + u

dv

dx+

du

dx

dv

dxδx ' du

dxv + u

dv

dx


The product rule


w ′ = (uv)′ = u′v + uv ′ dwdx

= ddx

(uv) = dudxv + u dv

dx.



δw = (δu) v + u (δv) + (δu)(δv)

δw

δx=δu

δxv + u

δv

δx+δu

δx

δv

δxδx

' du

dxv + u

dv

dx+

du

dx

dv

dxδx

' du

dxv + u

dv

dx


The product rule


w ′ = (uv)′ = u′v + uv ′ dwdx

= ddx

(uv) = dudxv + u dv

dx.



δw = (δu) v + u (δv) + (δu)(δv)

δw

δx=δu

δxv + u

δv

δx+δu

δx

δv

δxδx

' du

dxv + u

dv

dx+

du

dx

dv

dxδx ' du

dxv + u

dv

dx


The product rule


w ′ = (uv)′ = u′v + uv ′ dwdx

= ddx

(uv) = dudxv + u dv

dx.



δw = (δu) v + u (δv) + (δu)(δv)

δw

δx=δu

δxv + u

δv

δx+δu

δx

δv

δxδx

' du

dxv + u

dv

dx+

du

dx

dv

dxδx ' du

dxv + u

dv

dx


Examples of the product rule

(uv)′ = u′v + uv ′

d

dx(sin(x)cos(x)) = sin′(x)cos(x) + sin(x)cos′(x)

= cos(x)cos(x) + sin(x)(− sin(x))

= cos(x)2 − sin(x)2

d

dx(x3log(x)) = 3x2log(x) + x3log′(x)

= 3x2log(x) + x3(x−1)

= (3 log(x) + 1)x2

d

dx(eaxsin(bx)) = a eaxsin(bx) + eaxb cos(bx)

= eax(a sin(bx) + b cos(bx))


(uv)′ = u′v + uv ′

d




d


= 3x2log(x) + x3(x−1)

= (3 log(x) + 1)x2

d




(uv)′ = u′v + uv ′

d

dx(sin(x)cos(x))

= sin′(x)cos(x) + sin(x)cos′(x)



d


= 3x2log(x) + x3(x−1)

= (3 log(x) + 1)x2

d




(uv)′ = u′v + uv ′

d




d


= 3x2log(x) + x3(x−1)

= (3 log(x) + 1)x2

d




(uv)′ = u′v + uv ′

d




d


= 3x2log(x) + x3(x−1)

= (3 log(x) + 1)x2

d




(uv)′ = u′v + uv ′

d




d


= 3x2log(x) + x3(x−1)

= (3 log(x) + 1)x2

d




(uv)′ = u′v + uv ′

d




d

dx(x3log(x))

= 3x2log(x) + x3log′(x)

= 3x2log(x) + x3(x−1)

= (3 log(x) + 1)x2

d




(uv)′ = u′v + uv ′

d




d


= 3x2log(x) + x3(x−1)

= (3 log(x) + 1)x2

d




(uv)′ = u′v + uv ′

d




d


= 3x2log(x) + x3(x−1)

= (3 log(x) + 1)x2

d




(uv)′ = u′v + uv ′

d




d


= 3x2log(x) + x3(x−1)

= (3 log(x) + 1)x2

d




(uv)′ = u′v + uv ′

d




d


= 3x2log(x) + x3(x−1)

= (3 log(x) + 1)x2

d

dx(eaxsin(bx))

= a eaxsin(bx) + eaxb cos(bx)



(uv)′ = u′v + uv ′

d




d


= 3x2log(x) + x3(x−1)

= (3 log(x) + 1)x2

d




(uv)′ = u′v + uv ′

d




d


= 3x2log(x) + x3(x−1)

= (3 log(x) + 1)x2

d



The quotient rule

I Consider variables u and v depending on x , and put w = u/v . Then

w ′ =(uv

)′=

u′v − uv ′

v 2

I Indeed: u = vw

, so u′ = v ′w + vw ′ (product rule) , so

w ′ =u′ − v ′w

v

=u′

v− v ′.(u/v)

v=

u′

v− uv ′

v 2=

u′v − uv ′

v 2.

The quotient rule


w ′ =(uv

)′=

u′v − uv ′

v 2

I Indeed: u = vw


w ′ =u′ − v ′w

v

=u′

v− v ′.(u/v)

v=

u′

v− uv ′

v 2=

u′v − uv ′

v 2.

The quotient rule


w ′ =(uv

)′=

u′v − uv ′

v 2

I Indeed: u = vw


w ′ =u′ − v ′w

v

=u′

v− v ′.(u/v)

v=

u′

v− uv ′

v 2=

u′v − uv ′

v 2.

The quotient rule


w ′ =(uv

)′=

u′v − uv ′

v 2

I Indeed: u = vw , so u′ = v ′w + vw ′ (product rule)

, so

w ′ =u′ − v ′w

v

=u′

v− v ′.(u/v)

v=

u′

v− uv ′

v 2=

u′v − uv ′

v 2.

The quotient rule


w ′ =(uv

)′=

u′v − uv ′

v 2

I Indeed: u = vw , so u′ = v ′w + vw ′ (product rule) , so

w ′ =u′ − v ′w

v

=u′

v− v ′.(u/v)

v=

u′

v− uv ′

v 2=

u′v − uv ′

v 2.

The quotient rule


w ′ =(uv

)′=

u′v − uv ′

v 2


w ′ =u′ − v ′w

v=

u′

v− v ′.(u/v)

v

=u′

v− uv ′

v 2=

u′v − uv ′

v 2.

The quotient rule


w ′ =(uv

)′=

u′v − uv ′

v 2


w ′ =u′ − v ′w

v=

u′

v− v ′.(u/v)

v=

u′

v− uv ′

v 2

=u′v − uv ′

v 2.

The quotient rule


w ′ =(uv

)′=

u′v − uv ′

v 2


w ′ =u′ − v ′w

v=

u′

v− v ′.(u/v)

v=

u′

v− uv ′

v 2=

u′v − uv ′

v 2.

Examples of the quotient rule

d

dx

(x

log(x)

)=

1.log(x)− xx−1

log(x)2=

log(x)− 1

log(x)2= log(x)−1 − log(x)−2

(Aside: x/ log(x) ' ( number of primes ≤ x))

d

dx

(x

1− x2

)=

1.(1− x2)− x .(−2x)

(1− x2)2=

1− x2 + 2x2

(1− x2)2=

1 + x2

(1− x2)2

Now consider tan′(x), remembering that tan(x) = sin(x)/ cos(x).

d

dx

(sin(x)

cos(x)

)=

sin′(x)cos(x)− sin(x)cos′(x)

cos(x)2

=cos(x)cos(x)− sin(x)(− sin(x))

cos(x)2

=cos(x)2 + sin(x)2

cos(x)2=

1

cos(x)2= sec(x)2


d

dx

(x

log(x)

)

=1.log(x)− xx−1

log(x)2=

log(x)− 1



d

dx

(x

1− x2

)=

1.(1− x2)− x .(−2x)

(1− x2)2=

1− x2 + 2x2

(1− x2)2=

1 + x2

(1− x2)2


d

dx

(sin(x)

cos(x)

)=


cos(x)2


cos(x)2

=cos(x)2 + sin(x)2

cos(x)2=

1

cos(x)2= sec(x)2


d

dx

(x

log(x)

)=

1.log(x)− xx−1

log(x)2

=log(x)− 1



d

dx

(x

1− x2

)=

1.(1− x2)− x .(−2x)

(1− x2)2=

1− x2 + 2x2

(1− x2)2=

1 + x2

(1− x2)2


d

dx

(sin(x)

cos(x)

)=


cos(x)2


cos(x)2

=cos(x)2 + sin(x)2

cos(x)2=

1

cos(x)2= sec(x)2


d

dx

(x

log(x)

)=

1.log(x)− xx−1

log(x)2=

log(x)− 1

log(x)2

= log(x)−1 − log(x)−2


d

dx

(x

1− x2

)=

1.(1− x2)− x .(−2x)

(1− x2)2=

1− x2 + 2x2

(1− x2)2=

1 + x2

(1− x2)2


d

dx

(sin(x)

cos(x)

)=


cos(x)2


cos(x)2

=cos(x)2 + sin(x)2

cos(x)2=

1

cos(x)2= sec(x)2


d

dx

(x

log(x)

)=

1.log(x)− xx−1

log(x)2=

log(x)− 1



d

dx

(x

1− x2

)=

1.(1− x2)− x .(−2x)

(1− x2)2=

1− x2 + 2x2

(1− x2)2=

1 + x2

(1− x2)2


d

dx

(sin(x)

cos(x)

)=


cos(x)2


cos(x)2

=cos(x)2 + sin(x)2

cos(x)2=

1

cos(x)2= sec(x)2


d

dx

(x

log(x)

)=

1.log(x)− xx−1

log(x)2=

log(x)− 1



d

dx

(x

1− x2

)=

1.(1− x2)− x .(−2x)

(1− x2)2=

1− x2 + 2x2

(1− x2)2=

1 + x2

(1− x2)2


d

dx

(sin(x)

cos(x)

)=


cos(x)2


cos(x)2

=cos(x)2 + sin(x)2

cos(x)2=

1

cos(x)2= sec(x)2


d

dx

(x

log(x)

)=

1.log(x)− xx−1

log(x)2=

log(x)− 1



d

dx

(x

1− x2

)

=1.(1− x2)− x .(−2x)

(1− x2)2=

1− x2 + 2x2

(1− x2)2=

1 + x2

(1− x2)2


d

dx

(sin(x)

cos(x)

)=


cos(x)2


cos(x)2

=cos(x)2 + sin(x)2

cos(x)2=

1

cos(x)2= sec(x)2


d

dx

(x

log(x)

)=

1.log(x)− xx−1

log(x)2=

log(x)− 1



d

dx

(x

1− x2

)=

1.(1− x2)− x .(−2x)

(1− x2)2

=1− x2 + 2x2

(1− x2)2=

1 + x2

(1− x2)2


d

dx

(sin(x)

cos(x)

)=


cos(x)2


cos(x)2

=cos(x)2 + sin(x)2

cos(x)2=

1

cos(x)2= sec(x)2


d

dx

(x

log(x)

)=

1.log(x)− xx−1

log(x)2=

log(x)− 1



d

dx

(x

1− x2

)=

1.(1− x2)− x .(−2x)

(1− x2)2=

1− x2 + 2x2

(1− x2)2

=1 + x2

(1− x2)2


d

dx

(sin(x)

cos(x)

)=


cos(x)2


cos(x)2

=cos(x)2 + sin(x)2

cos(x)2=

1

cos(x)2= sec(x)2


d

dx

(x

log(x)

)=

1.log(x)− xx−1

log(x)2=

log(x)− 1



d

dx

(x

1− x2

)=

1.(1− x2)− x .(−2x)

(1− x2)2=

1− x2 + 2x2

(1− x2)2=

1 + x2

(1− x2)2


d

dx

(sin(x)

cos(x)

)=


cos(x)2


cos(x)2

=cos(x)2 + sin(x)2

cos(x)2=

1

cos(x)2= sec(x)2


d

dx

(x

log(x)

)=

1.log(x)− xx−1

log(x)2=

log(x)− 1



d

dx

(x

1− x2

)=

1.(1− x2)− x .(−2x)

(1− x2)2=

1− x2 + 2x2

(1− x2)2=

1 + x2

(1− x2)2


d

dx

(sin(x)

cos(x)

)=


cos(x)2


cos(x)2

=cos(x)2 + sin(x)2

cos(x)2=

1

cos(x)2= sec(x)2


d

dx

(x

log(x)

)=

1.log(x)− xx−1

log(x)2=

log(x)− 1



d

dx

(x

1− x2

)=

1.(1− x2)− x .(−2x)

(1− x2)2=

1− x2 + 2x2

(1− x2)2=

1 + x2

(1− x2)2


d

dx

(sin(x)

cos(x)

)

=sin′(x)cos(x)− sin(x)cos′(x)

cos(x)2


cos(x)2

=cos(x)2 + sin(x)2

cos(x)2=

1

cos(x)2= sec(x)2


d

dx

(x

log(x)

)=

1.log(x)− xx−1

log(x)2=

log(x)− 1



d

dx

(x

1− x2

)=

1.(1− x2)− x .(−2x)

(1− x2)2=

1− x2 + 2x2

(1− x2)2=

1 + x2

(1− x2)2


d

dx

(sin(x)

cos(x)

)=


cos(x)2


cos(x)2

=cos(x)2 + sin(x)2

cos(x)2=

1

cos(x)2= sec(x)2


d

dx

(x

log(x)

)=

1.log(x)− xx−1

log(x)2=

log(x)− 1



d

dx

(x

1− x2

)=

1.(1− x2)− x .(−2x)

(1− x2)2=

1− x2 + 2x2

(1− x2)2=

1 + x2

(1− x2)2


d

dx

(sin(x)

cos(x)

)=


cos(x)2


cos(x)2

=cos(x)2 + sin(x)2

cos(x)2=

1

cos(x)2= sec(x)2


d

dx

(x

log(x)

)=

1.log(x)− xx−1

log(x)2=

log(x)− 1



d

dx

(x

1− x2

)=

1.(1− x2)− x .(−2x)

(1− x2)2=

1− x2 + 2x2

(1− x2)2=

1 + x2

(1− x2)2


d

dx

(sin(x)

cos(x)

)=


cos(x)2


cos(x)2

=cos(x)2 + sin(x)2

cos(x)2

=1

cos(x)2= sec(x)2


d

dx

(x

log(x)

)=

1.log(x)− xx−1

log(x)2=

log(x)− 1



d

dx

(x

1− x2

)=

1.(1− x2)− x .(−2x)

(1− x2)2=

1− x2 + 2x2

(1− x2)2=

1 + x2

(1− x2)2


d

dx

(sin(x)

cos(x)

)=


cos(x)2


cos(x)2

=cos(x)2 + sin(x)2

cos(x)2=

1

cos(x)2

= sec(x)2


d

dx

(x

log(x)

)=

1.log(x)− xx−1

log(x)2=

log(x)− 1



d

dx

(x

1− x2

)=

1.(1− x2)− x .(−2x)

(1− x2)2=

1− x2 + 2x2

(1− x2)2=

1 + x2

(1− x2)2


d

dx

(sin(x)

cos(x)

)=


cos(x)2


cos(x)2

=cos(x)2 + sin(x)2

cos(x)2=

1

cos(x)2= sec(x)2

The chain rule

I Suppose that y depends on u, and u depends on x . Then

dy

dx=

dy

du

du

dx

I If x changes to x + δx , then u changes to u + δu and y changes toy + δy .

Clearlyδy

δx=δy

δu

δu

δx.

In the limit, δx , δu and δy all approach zero, and we get

dy

dx=

dy

du

du

dx.

I Alternative notation: suppose that f (x) = g(h(x)). Then

f ′(x) = g ′(h(x))h′(x)

The chain rule


dy

dx=

dy

du

du

dx


Clearlyδy

δx=δy

δu

δu

δx.


dy

dx=

dy

du

du

dx.


f ′(x) = g ′(h(x))h′(x)

The chain rule


dy

dx=

dy

du

du

dx


Clearlyδy

δx=δy

δu

δu

δx.


dy

dx=

dy

du

du

dx.


f ′(x) = g ′(h(x))h′(x)

The chain rule


dy

dx=

dy

du

du

dx

I If x changes to x + δx , then u changes to u + δu and y changes toy + δy . Clearly

δy

δx=δy

δu

δu

δx.


dy

dx=

dy

du

du

dx.


f ′(x) = g ′(h(x))h′(x)

The chain rule


dy

dx=

dy

du

du

dx


δy

δx=δy

δu

δu

δx.


dy

dx=

dy

du

du

dx.


f ′(x) = g ′(h(x))h′(x)

The chain rule


dy

dx=

dy

du

du

dx


δy

δx=δy

δu

δu

δx.


dy

dx=

dy

du

du

dx.


f ′(x) = g ′(h(x))h′(x)

Examples of the chain rule

I Consider y = cos(x2).

This is y = cos(u), where u = x2.

du

dx= 2x

dy

du= − sin(u) = − sin(x2)

dy

dx=

dy

du

du

dx= − sin(x2).2x = −2x sin(x2).

I Consider f (x) = exp(sin(x)).

f ′(x) = exp′(sin(x)). sin′(x) = exp(sin(x)) cos(x).

I Consider y = a sin(bx + c).

Put u = bx + c, so y = a sin(u).Then du

dx= b and dy

du= a cos(u) so

dy

dx=

dy

du

du

dx= a cos(u).b = ab cos(u) = ab cos(bx + c).


I Consider y = cos(x2).

This is y = cos(u), where u = x2.

du

dx= 2x

dy


dy

dx=

dy

du

du

dx= − sin(x2).2x = −2x sin(x2).





dx= b and dy

du= a cos(u) so

dy

dx=

dy

du

du



I Consider y = cos(x2). This is y = cos(u), where u = x2.

du

dx= 2x

dy


dy

dx=

dy

du

du

dx= − sin(x2).2x = −2x sin(x2).





dx= b and dy

du= a cos(u) so

dy

dx=

dy

du

du




du

dx= 2x

dy


dy

dx=

dy

du

du

dx= − sin(x2).2x = −2x sin(x2).





dx= b and dy

du= a cos(u) so

dy

dx=

dy

du

du




du

dx= 2x

dy

du= − sin(u)

= − sin(x2)

dy

dx=

dy

du

du

dx= − sin(x2).2x = −2x sin(x2).





dx= b and dy

du= a cos(u) so

dy

dx=

dy

du

du




du

dx= 2x

dy


dy

dx=

dy

du

du

dx= − sin(x2).2x = −2x sin(x2).





dx= b and dy

du= a cos(u) so

dy

dx=

dy

du

du




du

dx= 2x

dy


dy

dx=

dy

du

du

dx

= − sin(x2).2x = −2x sin(x2).





dx= b and dy

du= a cos(u) so

dy

dx=

dy

du

du




du

dx= 2x

dy


dy

dx=

dy

du

du

dx= − sin(x2).2x

= −2x sin(x2).





dx= b and dy

du= a cos(u) so

dy

dx=

dy

du

du




du

dx= 2x

dy


dy

dx=

dy

du

du

dx= − sin(x2).2x = −2x sin(x2).





dx= b and dy

du= a cos(u) so

dy

dx=

dy

du

du




du

dx= 2x

dy


dy

dx=

dy

du

du

dx= − sin(x2).2x = −2x sin(x2).





dx= b and dy

du= a cos(u) so

dy

dx=

dy

du

du




du

dx= 2x

dy


dy

dx=

dy

du

du

dx= − sin(x2).2x = −2x sin(x2).


f ′(x) = exp′(sin(x)). sin′(x)

= exp(sin(x)) cos(x).



dx= b and dy

du= a cos(u) so

dy

dx=

dy

du

du




du

dx= 2x

dy


dy

dx=

dy

du

du

dx= − sin(x2).2x = −2x sin(x2).





dx= b and dy

du= a cos(u) so

dy

dx=

dy

du

du




du

dx= 2x

dy


dy

dx=

dy

du

du

dx= − sin(x2).2x = −2x sin(x2).





dx= b and dy

du= a cos(u) so

dy

dx=

dy

du

du




du

dx= 2x

dy


dy

dx=

dy

du

du

dx= − sin(x2).2x = −2x sin(x2).



I Consider y = a sin(bx + c). Put u = bx + c, so y = a sin(u).

Then dudx

= b and dydu

= a cos(u) so

dy

dx=

dy

du

du




du

dx= 2x

dy


dy

dx=

dy

du

du

dx= − sin(x2).2x = −2x sin(x2).



I Consider y = a sin(bx + c). Put u = bx + c, so y = a sin(u).Then du

dx= b

and dydu

= a cos(u) so

dy

dx=

dy

du

du




du

dx= 2x

dy


dy

dx=

dy

du

du

dx= − sin(x2).2x = −2x sin(x2).




dx= b and dy

du= a cos(u)

so

dy

dx=

dy

du

du




du

dx= 2x

dy


dy

dx=

dy

du

du

dx= − sin(x2).2x = −2x sin(x2).




dx= b and dy

du= a cos(u) so

dy

dx=

dy

du

du

dx

= a cos(u).b = ab cos(u) = ab cos(bx + c).



du

dx= 2x

dy


dy

dx=

dy

du

du

dx= − sin(x2).2x = −2x sin(x2).




dx= b and dy

du= a cos(u) so

dy

dx=

dy

du

du

dx= a cos(u).b = ab cos(u)

= ab cos(bx + c).



du

dx= 2x

dy


dy

dx=

dy

du

du

dx= − sin(x2).2x = −2x sin(x2).




dx= b and dy

du= a cos(u) so

dy

dx=

dy

du

du


The power rule

I If u depends on x and n does not, then

d

dx(un) = nun−1 du

dx

I Reason: If y = un then dydu

= nun−1 so dydx

= dydu

dudx

= nun−1 dudx

I Consider y =√

1 + x2. This is y = u1/2, where u = 1 + x2. Then

dy

du=

1

2u−1/2 =

1

2√

1 + x2

du

dx= 2x

dy

dx=

1

2√

1 + x22x =

x√1 + x2

.

I ddx

(sin(x)5

)= 5 sin(x)4cos(x)

I ddx

(log(x)3

)= 3 log(x)2x−1 = 3 log(x)2/x

The power rule


d

dx(un) = nun−1 du

dx


= nun−1 so dydx

= dydu

dudx

= nun−1 dudx

I Consider y =√


dy

du=

1

2u−1/2 =

1

2√

1 + x2

du

dx= 2x

dy

dx=

1

2√

1 + x22x =

x√1 + x2

.

I ddx

(sin(x)5

)= 5 sin(x)4cos(x)

I ddx

(log(x)3

)= 3 log(x)2x−1 = 3 log(x)2/x

The power rule


d

dx(un) = nun−1 du

dx


= nun−1 so dydx

= dydu

dudx

= nun−1 dudx

I Consider y =√


dy

du=

1

2u−1/2 =

1

2√

1 + x2

du

dx= 2x

dy

dx=

1

2√

1 + x22x =

x√1 + x2

.

I ddx

(sin(x)5

)= 5 sin(x)4cos(x)

I ddx

(log(x)3

)= 3 log(x)2x−1 = 3 log(x)2/x

The power rule


d

dx(un) = nun−1 du

dx


= nun−1 so dydx

= dydu

dudx

= nun−1 dudx

I Consider y =√


dy

du=

1

2u−1/2 =

1

2√

1 + x2

du

dx= 2x

dy

dx=

1

2√

1 + x22x =

x√1 + x2

.

I ddx

(sin(x)5

)= 5 sin(x)4cos(x)

I ddx

(log(x)3

)= 3 log(x)2x−1 = 3 log(x)2/x

The power rule


d

dx(un) = nun−1 du

dx


= nun−1 so dydx

= dydu

dudx

= nun−1 dudx

I Consider y =√


dy

du=

1

2u−1/2 =

1

2√

1 + x2

du

dx= 2x

dy

dx=

1

2√

1 + x22x =

x√1 + x2

.

I ddx

(sin(x)5

)= 5 sin(x)4cos(x)

I ddx

(log(x)3

)= 3 log(x)2x−1 = 3 log(x)2/x

The power rule


d

dx(un) = nun−1 du

dx


= nun−1 so dydx

= dydu

dudx

= nun−1 dudx

I Consider y =√


dy

du=

1

2u−1/2 =

1

2√

1 + x2

du

dx= 2x

dy

dx=

1

2√

1 + x22x =

x√1 + x2

.

I ddx

(sin(x)5

)= 5 sin(x)4cos(x)

I ddx

(log(x)3

)= 3 log(x)2x−1 = 3 log(x)2/x

The power rule


d

dx(un) = nun−1 du

dx


= nun−1 so dydx

= dydu

dudx

= nun−1 dudx

I Consider y =√


dy

du=

1

2u−1/2 =

1

2√

1 + x2

du

dx= 2x

dy

dx=

1

2√

1 + x22x =

x√1 + x2

.

I ddx

(sin(x)5

)= 5 sin(x)4cos(x)

I ddx

(log(x)3

)= 3 log(x)2x−1 = 3 log(x)2/x

The power rule


d

dx(un) = nun−1 du

dx


= nun−1 so dydx

= dydu

dudx

= nun−1 dudx

I Consider y =√


dy

du=

1

2u−1/2 =

1

2√

1 + x2

du

dx= 2x

dy

dx=

1

2√

1 + x22x =

x√1 + x2

.

I ddx

(sin(x)5

)= 5 sin(x)4cos(x)

I ddx

(log(x)3

)= 3 log(x)2x−1 = 3 log(x)2/x

The logarithmic rule

I . d

dxlog(u) =

1

u

du

dx

du

dx= u

d

dxlog(u)

I .d

dxlog(cos(x))

=1

cos(x)cos′(x) =

− sin(x)

cos(x)= − tan(x)

I . d

dxlog(1 + x2)

=ddx

(1 + x2)

1 + x2=

2x

1 + x2

I Consider y = xx

, so log(y) = x log(x). Then

d

dxlog(y) =

d

dx(x log(x))

= 1. log(x) + x .x−1 = log(x) + 1

dy

dx= y

d

dxlog(y)

= xx(log(x) + 1).


I . d

dxlog(u) =

1

u

du

dx

du

dx= u

d

dxlog(u)

I .d

dxlog(cos(x))

=1

cos(x)cos′(x) =

− sin(x)

cos(x)= − tan(x)

I . d

dxlog(1 + x2)

=ddx

(1 + x2)

1 + x2=

2x

1 + x2

I Consider y = xx


d

dxlog(y) =

d

dx(x log(x))

= 1. log(x) + x .x−1 = log(x) + 1

dy

dx= y

d

dxlog(y)

= xx(log(x) + 1).


I . d

dxlog(u) =

1

u

du

dx

du

dx= u

d

dxlog(u)

I .d

dxlog(cos(x))

=1

cos(x)cos′(x) =

− sin(x)

cos(x)= − tan(x)

I . d

dxlog(1 + x2)

=ddx

(1 + x2)

1 + x2=

2x

1 + x2

I Consider y = xx


d

dxlog(y) =

d

dx(x log(x))

= 1. log(x) + x .x−1 = log(x) + 1

dy

dx= y

d

dxlog(y)

= xx(log(x) + 1).


I . d

dxlog(u) =

1

u

du

dx

du

dx= u

d

dxlog(u)

I .d

dxlog(cos(x))

=1

cos(x)cos′(x) =

− sin(x)

cos(x)= − tan(x)

I . d

dxlog(1 + x2)

=ddx

(1 + x2)

1 + x2=

2x

1 + x2

I Consider y = xx


d

dxlog(y) =

d

dx(x log(x))

= 1. log(x) + x .x−1 = log(x) + 1

dy

dx= y

d

dxlog(y)

= xx(log(x) + 1).


I . d

dxlog(u) =

1

u

du

dx

du

dx= u

d

dxlog(u)

I .d

dxlog(cos(x)) =

1

cos(x)cos′(x)

=− sin(x)

cos(x)= − tan(x)

I . d

dxlog(1 + x2)

=ddx

(1 + x2)

1 + x2=

2x

1 + x2

I Consider y = xx


d

dxlog(y) =

d

dx(x log(x))

= 1. log(x) + x .x−1 = log(x) + 1

dy

dx= y

d

dxlog(y)

= xx(log(x) + 1).


I . d

dxlog(u) =

1

u

du

dx

du

dx= u

d

dxlog(u)

I .d

dxlog(cos(x)) =

1

cos(x)cos′(x) =

− sin(x)

cos(x)

= − tan(x)

I . d

dxlog(1 + x2)

=ddx

(1 + x2)

1 + x2=

2x

1 + x2

I Consider y = xx


d

dxlog(y) =

d

dx(x log(x))

= 1. log(x) + x .x−1 = log(x) + 1

dy

dx= y

d

dxlog(y)

= xx(log(x) + 1).


I . d

dxlog(u) =

1

u

du

dx

du

dx= u

d

dxlog(u)

I .d

dxlog(cos(x)) =

1

cos(x)cos′(x) =

− sin(x)

cos(x)= − tan(x)

I . d

dxlog(1 + x2)

=ddx

(1 + x2)

1 + x2=

2x

1 + x2

I Consider y = xx


d

dxlog(y) =

d

dx(x log(x))

= 1. log(x) + x .x−1 = log(x) + 1

dy

dx= y

d

dxlog(y)

= xx(log(x) + 1).


I . d

dxlog(u) =

1

u

du

dx

du

dx= u

d

dxlog(u)

I .d

dxlog(cos(x)) =

1

cos(x)cos′(x) =

− sin(x)

cos(x)= − tan(x)

I . d

dxlog(1 + x2)

=ddx

(1 + x2)

1 + x2=

2x

1 + x2

I Consider y = xx


d

dxlog(y) =

d

dx(x log(x))

= 1. log(x) + x .x−1 = log(x) + 1

dy

dx= y

d

dxlog(y)

= xx(log(x) + 1).


I . d

dxlog(u) =

1

u

du

dx

du

dx= u

d

dxlog(u)

I .d

dxlog(cos(x)) =

1

cos(x)cos′(x) =

− sin(x)

cos(x)= − tan(x)

I . d

dxlog(1 + x2) =

ddx

(1 + x2)

1 + x2

=2x

1 + x2

I Consider y = xx


d

dxlog(y) =

d

dx(x log(x))

= 1. log(x) + x .x−1 = log(x) + 1

dy

dx= y

d

dxlog(y)

= xx(log(x) + 1).


I . d

dxlog(u) =

1

u

du

dx

du

dx= u

d

dxlog(u)

I .d

dxlog(cos(x)) =

1

cos(x)cos′(x) =

− sin(x)

cos(x)= − tan(x)

I . d

dxlog(1 + x2) =

ddx

(1 + x2)

1 + x2=

2x

1 + x2

I Consider y = xx


d

dxlog(y) =

d

dx(x log(x))

= 1. log(x) + x .x−1 = log(x) + 1

dy

dx= y

d

dxlog(y)

= xx(log(x) + 1).


I . d

dxlog(u) =

1

u

du

dx

du

dx= u

d

dxlog(u)

I .d

dxlog(cos(x)) =

1

cos(x)cos′(x) =

− sin(x)

cos(x)= − tan(x)

I . d

dxlog(1 + x2) =

ddx

(1 + x2)

1 + x2=

2x

1 + x2

I Consider y = xx


d

dxlog(y) =

d

dx(x log(x))

= 1. log(x) + x .x−1 = log(x) + 1

dy

dx= y

d

dxlog(y)

= xx(log(x) + 1).


I . d

dxlog(u) =

1

u

du

dx

du

dx= u

d

dxlog(u)

I .d

dxlog(cos(x)) =

1

cos(x)cos′(x) =

− sin(x)

cos(x)= − tan(x)

I . d

dxlog(1 + x2) =

ddx

(1 + x2)

1 + x2=

2x

1 + x2

I Consider y = xx , so log(y) = x log(x).

Then

d

dxlog(y) =

d

dx(x log(x))

= 1. log(x) + x .x−1 = log(x) + 1

dy

dx= y

d

dxlog(y)

= xx(log(x) + 1).


I . d

dxlog(u) =

1

u

du

dx

du

dx= u

d

dxlog(u)

I .d

dxlog(cos(x)) =

1

cos(x)cos′(x) =

− sin(x)

cos(x)= − tan(x)

I . d

dxlog(1 + x2) =

ddx

(1 + x2)

1 + x2=

2x

1 + x2

I Consider y = xx , so log(y) = x log(x). Then

d

dxlog(y) =

d

dx(x log(x))

= 1. log(x) + x .x−1 = log(x) + 1

dy

dx= y

d

dxlog(y)

= xx(log(x) + 1).


I . d

dxlog(u) =

1

u

du

dx

du

dx= u

d

dxlog(u)

I .d

dxlog(cos(x)) =

1

cos(x)cos′(x) =

− sin(x)

cos(x)= − tan(x)

I . d

dxlog(1 + x2) =

ddx

(1 + x2)

1 + x2=

2x

1 + x2


d

dxlog(y) =

d

dx(x log(x))

= 1. log(x) + x .x−1

= log(x) + 1

dy

dx= y

d

dxlog(y)

= xx(log(x) + 1).


I . d

dxlog(u) =

1

u

du

dx

du

dx= u

d

dxlog(u)

I .d

dxlog(cos(x)) =

1

cos(x)cos′(x) =

− sin(x)

cos(x)= − tan(x)

I . d

dxlog(1 + x2) =

ddx

(1 + x2)

1 + x2=

2x

1 + x2


d

dxlog(y) =

d

dx(x log(x))

= 1. log(x) + x .x−1 = log(x) + 1

dy

dx= y

d

dxlog(y)

= xx(log(x) + 1).


I . d

dxlog(u) =

1

u

du

dx

du

dx= u

d

dxlog(u)

I .d

dxlog(cos(x)) =

1

cos(x)cos′(x) =

− sin(x)

cos(x)= − tan(x)

I . d

dxlog(1 + x2) =

ddx

(1 + x2)

1 + x2=

2x

1 + x2


d

dxlog(y) =

d

dx(x log(x))

= 1. log(x) + x .x−1 = log(x) + 1

dy

dx= y

d

dxlog(y)

= xx(log(x) + 1).


I . d

dxlog(u) =

1

u

du

dx

du

dx= u

d

dxlog(u)

I .d

dxlog(cos(x)) =

1

cos(x)cos′(x) =

− sin(x)

cos(x)= − tan(x)

I . d

dxlog(1 + x2) =

ddx

(1 + x2)

1 + x2=

2x

1 + x2


d

dxlog(y) =

d

dx(x log(x))

= 1. log(x) + x .x−1 = log(x) + 1

dy

dx= y

d

dxlog(y)

= xx(log(x) + 1).

The inverse function rule

I If x and y are interdependent variables, then

dx

dy= 1/

dy

dx

I (Take limits in the obvious relation δxδy

= 1/ δyδx

.)

I Consider y = log(x)

, so x = ey .

dx

dy= ey = x

dy

dx= 1/

dx

dy=

1

x

I Alternative notation: if y = g(x) then x = f (y), where f = g−1 andg = f −1. Then

g ′(x) = 1/f ′(g(x))

I log′(x) = 1/ exp′(log(x)) = 1/ exp(log(x)) = 1/x .



dx

dy= 1/

dy

dx


= 1/ δyδx

.)


, so x = ey .

dx

dy= ey = x

dy

dx= 1/

dx

dy=

1

x


g ′(x) = 1/f ′(g(x))




dx

dy= 1/

dy

dx


= 1/ δyδx

.)


, so x = ey .

dx

dy= ey = x

dy

dx= 1/

dx

dy=

1

x


g ′(x) = 1/f ′(g(x))




dx

dy= 1/

dy

dx


= 1/ δyδx

.)


, so x = ey .

dx

dy= ey = x

dy

dx= 1/

dx

dy=

1

x


g ′(x) = 1/f ′(g(x))




dx

dy= 1/

dy

dx


= 1/ δyδx

.)

I Consider y = log(x), so x = ey .

dx

dy= ey = x

dy

dx= 1/

dx

dy=

1

x


g ′(x) = 1/f ′(g(x))




dx

dy= 1/

dy

dx


= 1/ δyδx

.)


dx

dy= ey = x

dy

dx= 1/

dx

dy=

1

x


g ′(x) = 1/f ′(g(x))




dx

dy= 1/

dy

dx


= 1/ δyδx

.)


dx

dy= ey = x

dy

dx= 1/

dx

dy

=1

x


g ′(x) = 1/f ′(g(x))




dx

dy= 1/

dy

dx


= 1/ δyδx

.)


dx

dy= ey = x

dy

dx= 1/

dx

dy=

1

x


g ′(x) = 1/f ′(g(x))




dx

dy= 1/

dy

dx


= 1/ δyδx

.)


dx

dy= ey = x

dy

dx= 1/

dx

dy=

1

x


g ′(x) = 1/f ′(g(x))




dx

dy= 1/

dy

dx


= 1/ δyδx

.)


dx

dy= ey = x

dy

dx= 1/

dx

dy=

1

x


g ′(x) = 1/f ′(g(x))


The arcsin function

I Consider y = arcsin(x)

, so x = sin(y).

dx

dy= sin′(y) = cos(y)

dy

dx= 1/

dx

dy= cos(y)−1.

I Also sin(y)2 + cos(y)2 = 1

, so

cos(y) =√

1− sin(y)2 =√

1− x2

cos(y)−1 = (1− x2)−1/2

I So arcsin′(x) = dydx

= (1− x2)−1/2.

The arcsin function

I Consider y = arcsin(x)

, so x = sin(y).

dx


dy

dx= 1/

dx

dy= cos(y)−1.


, so

cos(y) =√

1− sin(y)2 =√

1− x2

cos(y)−1 = (1− x2)−1/2


= (1− x2)−1/2.

The arcsin function

I Consider y = arcsin(x), so x = sin(y).

dx


dy

dx= 1/

dx

dy= cos(y)−1.


, so

cos(y) =√

1− sin(y)2 =√

1− x2

cos(y)−1 = (1− x2)−1/2


= (1− x2)−1/2.

The arcsin function


dx

dy

= sin′(y) = cos(y)

dy

dx= 1/

dx

dy= cos(y)−1.


, so

cos(y) =√

1− sin(y)2 =√

1− x2

cos(y)−1 = (1− x2)−1/2


= (1− x2)−1/2.

The arcsin function


dx


dy

dx= 1/

dx

dy= cos(y)−1.


, so

cos(y) =√

1− sin(y)2 =√

1− x2

cos(y)−1 = (1− x2)−1/2


= (1− x2)−1/2.

The arcsin function


dx


dy

dx= 1/

dx

dy= cos(y)−1.


, so

cos(y) =√

1− sin(y)2 =√

1− x2

cos(y)−1 = (1− x2)−1/2


= (1− x2)−1/2.

The arcsin function


dx


dy

dx= 1/

dx

dy= cos(y)−1.


, so

cos(y) =√

1− sin(y)2 =√

1− x2

cos(y)−1 = (1− x2)−1/2


= (1− x2)−1/2.

The arcsin function


dx


dy

dx= 1/

dx

dy= cos(y)−1.

I Also sin(y)2 + cos(y)2 = 1, so

cos(y) =√

1− sin(y)2

=√

1− x2

cos(y)−1 = (1− x2)−1/2


= (1− x2)−1/2.

The arcsin function


dx


dy

dx= 1/

dx

dy= cos(y)−1.


cos(y) =√

1− sin(y)2 =√

1− x2

cos(y)−1 = (1− x2)−1/2


= (1− x2)−1/2.

The arcsin function


dx


dy

dx= 1/

dx

dy= cos(y)−1.


cos(y) =√

1− sin(y)2 =√

1− x2

cos(y)−1 = (1− x2)−1/2


= (1− x2)−1/2.

The arcsin function


dx


dy

dx= 1/

dx

dy= cos(y)−1.


cos(y) =√

1− sin(y)2 =√

1− x2

cos(y)−1 = (1− x2)−1/2


= (1− x2)−1/2.

The arctanh function

I Consider y = arctanh(x)

, so x = tanh(y) = sinh(y)cosh(y)

.

dx

dy= tanh′(y)

=sinh′(y) cosh(y)− sinh(y) cosh′(y)

cosh(y)2

=cosh(y)2 − sinh(y)2

cosh(y)2

= 1− tanh(y)2 = 1− x2

dy

dx= 1/

dx

dy=

1

1− x2.

I So arctanh′(x) = dydx

= (1− x2)−1.


I Consider y = arctanh(x)

, so x = tanh(y) = sinh(y)cosh(y)

.

dx

dy= tanh′(y)


cosh(y)2


cosh(y)2

= 1− tanh(y)2 = 1− x2

dy

dx= 1/

dx

dy=

1

1− x2.


= (1− x2)−1.


I Consider y = arctanh(x), so x = tanh(y) = sinh(y)cosh(y)

.

dx

dy= tanh′(y)


cosh(y)2


cosh(y)2

= 1− tanh(y)2 = 1− x2

dy

dx= 1/

dx

dy=

1

1− x2.


= (1− x2)−1.



.

dx

dy= tanh′(y)


cosh(y)2


cosh(y)2

= 1− tanh(y)2 = 1− x2

dy

dx= 1/

dx

dy=

1

1− x2.


= (1− x2)−1.



.

dx

dy= tanh′(y)


cosh(y)2


cosh(y)2

= 1− tanh(y)2 = 1− x2

dy

dx= 1/

dx

dy=

1

1− x2.


= (1− x2)−1.



.

dx

dy= tanh′(y)


cosh(y)2


cosh(y)2

= 1− tanh(y)2 = 1− x2

dy

dx= 1/

dx

dy=

1

1− x2.


= (1− x2)−1.



.

dx

dy= tanh′(y)


cosh(y)2


cosh(y)2

= 1− tanh(y)2

= 1− x2

dy

dx= 1/

dx

dy=

1

1− x2.


= (1− x2)−1.



.

dx

dy= tanh′(y)


cosh(y)2


cosh(y)2

= 1− tanh(y)2 = 1− x2

dy

dx= 1/

dx

dy=

1

1− x2.


= (1− x2)−1.



.

dx

dy= tanh′(y)


cosh(y)2


cosh(y)2

= 1− tanh(y)2 = 1− x2

dy

dx= 1/

dx

dy=

1

1− x2.


= (1− x2)−1.



.

dx

dy= tanh′(y)


cosh(y)2


cosh(y)2

= 1− tanh(y)2 = 1− x2

dy

dx= 1/

dx

dy=

1

1− x2.


= (1− x2)−1.

Classes of functions

I If f (x) is a polynomial, then so is f ′(x).

I Eg f (x) = x + x10 + x100; f ′(x) = 1 + 10x9 + 100x99

I Eg f (x) = (x − 1)4 + (x + 1)4; f ′(x) = 4(x − 1)3 + 4(x + 1)3

I If f (x) is a rational function, then so is f ′(x).

I Eg f (x) = x2−1x2+1

; f ′(x) = 4x(x2+1)2

I Eg f (x) = 1x

+ 1x+1

+ 1x+2

; f ′(x) = − 1x2 − 1

(x+1)2 − 1(x+2)2

I If f (x) is a trigonometric polynomial, so is f ′(x).

I Eg f (x) = sin(x) + sin(3x)/3 + sin(5x)/5;f ′(x) = cos(x) + cos(3x) + cos(5x).

I Eg f (x) = sin(3x) + cos(3x); f ′(x) = 3 cos(3x)− 3 sin(3x).

I If f (x) is a polynomial times ex , so is f ′(x).

I Eg f (x) = (x + x2)ex ; f ′(x) = (1 + 3x + x2)ex .I Eg f (x) = (x4 − 4x3 + 12x2 − 24x + 24)ex ; f ′(x) = x4ex .


I If f (x) is a polynomial, then so is f ′(x).

I Eg f (x) = x + x10 + x100; f ′(x) = 1 + 10x9 + 100x99

I Eg f (x) = (x − 1)4 + (x + 1)4; f ′(x) = 4(x − 1)3 + 4(x + 1)3


I Eg f (x) = x2−1x2+1

; f ′(x) = 4x(x2+1)2

I Eg f (x) = 1x

+ 1x+1

+ 1x+2

; f ′(x) = − 1x2 − 1

(x+1)2 − 1(x+2)2







I If f (x) is a polynomial, then so is f ′(x).I Eg f (x) = x + x10 + x100; f ′(x) = 1 + 10x9 + 100x99

I Eg f (x) = (x − 1)4 + (x + 1)4; f ′(x) = 4(x − 1)3 + 4(x + 1)3


I Eg f (x) = x2−1x2+1

; f ′(x) = 4x(x2+1)2

I Eg f (x) = 1x

+ 1x+1

+ 1x+2

; f ′(x) = − 1x2 − 1

(x+1)2 − 1(x+2)2








I Eg f (x) = (x − 1)4 + (x + 1)4; f ′(x) = 4(x − 1)3 + 4(x + 1)3


I Eg f (x) = x2−1x2+1

; f ′(x) = 4x(x2+1)2

I Eg f (x) = 1x

+ 1x+1

+ 1x+2

; f ′(x) = − 1x2 − 1

(x+1)2 − 1(x+2)2








I Eg f (x) = (x − 1)4 + (x + 1)4; f ′(x) = 4(x − 1)3 + 4(x + 1)3


I Eg f (x) = x2−1x2+1

; f ′(x) = 4x(x2+1)2

I Eg f (x) = 1x

+ 1x+1

+ 1x+2

; f ′(x) = − 1x2 − 1

(x+1)2 − 1(x+2)2








I Eg f (x) = (x − 1)4 + (x + 1)4; f ′(x) = 4(x − 1)3 + 4(x + 1)3


I Eg f (x) = x2−1x2+1

; f ′(x) = 4x(x2+1)2

I Eg f (x) = 1x

+ 1x+1

+ 1x+2

; f ′(x) = − 1x2 − 1

(x+1)2 − 1(x+2)2








I Eg f (x) = (x − 1)4 + (x + 1)4; f ′(x) = 4(x − 1)3 + 4(x + 1)3


I Eg f (x) = x2−1x2+1

; f ′(x) = 4x(x2+1)2

I Eg f (x) = 1x

+ 1x+1

+ 1x+2

; f ′(x) = − 1x2 − 1

(x+1)2 − 1(x+2)2








I Eg f (x) = (x − 1)4 + (x + 1)4; f ′(x) = 4(x − 1)3 + 4(x + 1)3


I Eg f (x) = x2−1x2+1

; f ′(x) = 4x(x2+1)2

I Eg f (x) = 1x

+ 1x+1

+ 1x+2

; f ′(x) = − 1x2 − 1

(x+1)2 − 1(x+2)2








I Eg f (x) = (x − 1)4 + (x + 1)4; f ′(x) = 4(x − 1)3 + 4(x + 1)3


I Eg f (x) = x2−1x2+1

; f ′(x) = 4x(x2+1)2

I Eg f (x) = 1x

+ 1x+1

+ 1x+2

; f ′(x) = − 1x2 − 1

(x+1)2 − 1(x+2)2

I If f (x) is a trigonometric polynomial, so is f ′(x).I Eg f (x) = sin(x) + sin(3x)/3 + sin(5x)/5;

f ′(x) = cos(x) + cos(3x) + cos(5x).






I Eg f (x) = (x − 1)4 + (x + 1)4; f ′(x) = 4(x − 1)3 + 4(x + 1)3


I Eg f (x) = x2−1x2+1

; f ′(x) = 4x(x2+1)2

I Eg f (x) = 1x

+ 1x+1

+ 1x+2

; f ′(x) = − 1x2 − 1

(x+1)2 − 1(x+2)2


f ′(x) = cos(x) + cos(3x) + cos(5x).I Eg f (x) = sin(3x) + cos(3x); f ′(x) = 3 cos(3x)− 3 sin(3x).





I Eg f (x) = (x − 1)4 + (x + 1)4; f ′(x) = 4(x − 1)3 + 4(x + 1)3


I Eg f (x) = x2−1x2+1

; f ′(x) = 4x(x2+1)2

I Eg f (x) = 1x

+ 1x+1

+ 1x+2

; f ′(x) = − 1x2 − 1

(x+1)2 − 1(x+2)2







I Eg f (x) = (x − 1)4 + (x + 1)4; f ′(x) = 4(x − 1)3 + 4(x + 1)3


I Eg f (x) = x2−1x2+1

; f ′(x) = 4x(x2+1)2

I Eg f (x) = 1x

+ 1x+1

+ 1x+2

; f ′(x) = − 1x2 − 1

(x+1)2 − 1(x+2)2



I If f (x) is a polynomial times ex , so is f ′(x).I Eg f (x) = (x + x2)ex ; f ′(x) = (1 + 3x + x2)ex .

I Eg f (x) = (x4 − 4x3 + 12x2 − 24x + 24)ex ; f ′(x) = x4ex .



I Eg f (x) = (x − 1)4 + (x + 1)4; f ′(x) = 4(x − 1)3 + 4(x + 1)3


I Eg f (x) = x2−1x2+1

; f ′(x) = 4x(x2+1)2

I Eg f (x) = 1x

+ 1x+1

+ 1x+2

; f ′(x) = − 1x2 − 1

(x+1)2 − 1(x+2)2



I If f (x) is a polynomial times ex , so is f ′(x).I Eg f (x) = (x + x2)ex ; f ′(x) = (1 + 3x + x2)ex .I Eg f (x) = (x4 − 4x3 + 12x2 − 24x + 24)ex ; f ′(x) = x4ex .

Implicit differentiation

I Suppose that x and y are related by an equation such as y 4 + xy = x3.We cannot write y as a function of x , but we can still find dy/dx .

I Differentiate both sides. Terms in the equation involving y give terms inthe derivative involving dy/dx . Rearranging gives dy/dx in terms of x andy .

I Suppose that y 4 + xy = x3

, so

ddx

(y 4 + xy

)= d

dx

(x3)

= 3x2.

Also ddx

(y 4) = 4y 3 dydx

by the power rule

and ddx

(xy) = dxdxy + x dy

dx= y + x dy

dxby the product rule ; so

4y 3 dydx

+ y + x dydx

= 3x2

(4y 3 + x) dydx

= 3x2 − y

dydx

= 3x2−y4y3+x

.





, so

ddx

(y 4 + xy

)= d

dx

(x3)

= 3x2.

Also ddx

(y 4) = 4y 3 dydx

by the power rule

and ddx

(xy) = dxdxy + x dy

dx= y + x dy


4y 3 dydx

+ y + x dydx

= 3x2

(4y 3 + x) dydx

= 3x2 − y

dydx

= 3x2−y4y3+x

.





, so

ddx

(y 4 + xy

)= d

dx

(x3)

= 3x2.

Also ddx

(y 4) = 4y 3 dydx

by the power rule

and ddx

(xy) = dxdxy + x dy

dx= y + x dy


4y 3 dydx

+ y + x dydx

= 3x2

(4y 3 + x) dydx

= 3x2 − y

dydx

= 3x2−y4y3+x

.





, so

ddx

(y 4 + xy

)= d

dx

(x3)

= 3x2.

Also ddx

(y 4) = 4y 3 dydx

by the power rule

and ddx

(xy) = dxdxy + x dy

dx= y + x dy


4y 3 dydx

+ y + x dydx

= 3x2

(4y 3 + x) dydx

= 3x2 − y

dydx

= 3x2−y4y3+x

.




I Suppose that y 4 + xy = x3, so

ddx

(y 4 + xy

)= d

dx

(x3)

= 3x2.

Also ddx

(y 4) = 4y 3 dydx

by the power rule

and ddx

(xy) = dxdxy + x dy

dx= y + x dy


4y 3 dydx

+ y + x dydx

= 3x2

(4y 3 + x) dydx

= 3x2 − y

dydx

= 3x2−y4y3+x

.





ddx

(y 4 + xy

)= d

dx

(x3)

= 3x2.

Also ddx

(y 4) = 4y 3 dydx

by the power rule

and ddx

(xy) = dxdxy + x dy

dx= y + x dy


4y 3 dydx

+ y + x dydx

= 3x2

(4y 3 + x) dydx

= 3x2 − y

dydx

= 3x2−y4y3+x

.





ddx

(y 4 + xy

)= d

dx

(x3)

= 3x2.

Also ddx

(y 4) = 4y 3 dydx

by the power rule

and ddx

(xy) = dxdxy + x dy

dx= y + x dy

dxby the product rule

; so

4y 3 dydx

+ y + x dydx

= 3x2

(4y 3 + x) dydx

= 3x2 − y

dydx

= 3x2−y4y3+x

.





ddx

(y 4 + xy

)= d

dx

(x3)

= 3x2.

Also ddx

(y 4) = 4y 3 dydx

by the power rule

and ddx

(xy) = dxdxy + x dy

dx= y + x dy


4y 3 dydx

+ y + x dydx

= 3x2

(4y 3 + x) dydx

= 3x2 − y

dydx

= 3x2−y4y3+x

.





ddx

(y 4 + xy

)= d

dx

(x3)

= 3x2.

Also ddx

(y 4) = 4y 3 dydx

by the power rule

and ddx

(xy) = dxdxy + x dy

dx= y + x dy


4y 3 dydx

+ y + x dydx

= 3x2

(4y 3 + x) dydx

= 3x2 − y

dydx

= 3x2−y4y3+x

.





ddx

(y 4 + xy

)= d

dx

(x3)

= 3x2.

Also ddx

(y 4) = 4y 3 dydx

by the power rule

and ddx

(xy) = dxdxy + x dy

dx= y + x dy


4y 3 dydx

+ y + x dydx

= 3x2

(4y 3 + x) dydx

= 3x2 − y

dydx

= 3x2−y4y3+x

.

Implicit examples

I Suppose x + sin(x) = y − cos(y).

ddx

(x + sin(x)) = ddx

(y − cos(y))

1 + cos(x) = dydx

+ sin(y) dydx

dydx

= 1+cos(x)1+sin(y)

I Suppose y = exp(x2 + y 2).

dydx

= ddx

exp(x2 + y 2) = ddx

(ex2

ey2

)

= 2xex2

ey2

+ ex2

.2y dydxey

2

= 2(x + y dydx

) exp(x2 + y 2)

(1− 2y exp(x2 + y 2)) dydx

= 2x exp(x2 + y 2)

dydx

= 2x exp(x2+y2)

1−2y exp(x2+y2)

Implicit examples


ddx

(x + sin(x)) = ddx

(y − cos(y))

1 + cos(x) = dydx

+ sin(y) dydx

dydx

= 1+cos(x)1+sin(y)


dydx

= ddx

exp(x2 + y 2) = ddx

(ex2

ey2

)

= 2xex2

ey2

+ ex2

.2y dydxey

2

= 2(x + y dydx

) exp(x2 + y 2)

(1− 2y exp(x2 + y 2)) dydx

= 2x exp(x2 + y 2)

dydx

= 2x exp(x2+y2)

1−2y exp(x2+y2)

Implicit examples


ddx

(x + sin(x)) = ddx

(y − cos(y))

1 + cos(x) = dydx

+ sin(y) dydx

dydx

= 1+cos(x)1+sin(y)


dydx

= ddx

exp(x2 + y 2) = ddx

(ex2

ey2

)

= 2xex2

ey2

+ ex2

.2y dydxey

2

= 2(x + y dydx

) exp(x2 + y 2)

(1− 2y exp(x2 + y 2)) dydx

= 2x exp(x2 + y 2)

dydx

= 2x exp(x2+y2)

1−2y exp(x2+y2)

Implicit examples


ddx

(x + sin(x)) = ddx

(y − cos(y))

1 + cos(x) = dydx

+ sin(y) dydx

dydx

= 1+cos(x)1+sin(y)


dydx

= ddx

exp(x2 + y 2) = ddx

(ex2

ey2

)

= 2xex2

ey2

+ ex2

.2y dydxey

2

= 2(x + y dydx

) exp(x2 + y 2)

(1− 2y exp(x2 + y 2)) dydx

= 2x exp(x2 + y 2)

dydx

= 2x exp(x2+y2)

1−2y exp(x2+y2)

Implicit examples


ddx

(x + sin(x)) = ddx

(y − cos(y))

1 + cos(x) = dydx

+ sin(y) dydx

dydx

= 1+cos(x)1+sin(y)


dydx

= ddx

exp(x2 + y 2) = ddx

(ex2

ey2

)

= 2xex2

ey2

+ ex2

.2y dydxey

2

= 2(x + y dydx

) exp(x2 + y 2)

(1− 2y exp(x2 + y 2)) dydx

= 2x exp(x2 + y 2)

dydx

= 2x exp(x2+y2)

1−2y exp(x2+y2)

Implicit examples


ddx

(x + sin(x)) = ddx

(y − cos(y))

1 + cos(x) = dydx

+ sin(y) dydx

dydx

= 1+cos(x)1+sin(y)


dydx

= ddx

exp(x2 + y 2) = ddx

(ex2

ey2

)

= 2xex2

ey2

+ ex2

.2y dydxey

2

= 2(x + y dydx

) exp(x2 + y 2)

(1− 2y exp(x2 + y 2)) dydx

= 2x exp(x2 + y 2)

dydx

= 2x exp(x2+y2)

1−2y exp(x2+y2)

Implicit examples


ddx

(x + sin(x)) = ddx

(y − cos(y))

1 + cos(x) = dydx

+ sin(y) dydx

dydx

= 1+cos(x)1+sin(y)


dydx

= ddx

exp(x2 + y 2) = ddx

(ex2

ey2

)

= 2xex2

ey2

+ ex2

.2y dydxey

2

= 2(x + y dydx

) exp(x2 + y 2)

(1− 2y exp(x2 + y 2)) dydx

= 2x exp(x2 + y 2)

dydx

= 2x exp(x2+y2)

1−2y exp(x2+y2)

Implicit examples


ddx

(x + sin(x)) = ddx

(y − cos(y))

1 + cos(x) = dydx

+ sin(y) dydx

dydx

= 1+cos(x)1+sin(y)


dydx

= ddx

exp(x2 + y 2) = ddx

(ex2

ey2

)

= 2xex2

ey2

+ ex2

.2y dydxey

2

= 2(x + y dydx

) exp(x2 + y 2)

(1− 2y exp(x2 + y 2)) dydx

= 2x exp(x2 + y 2)

dydx

= 2x exp(x2+y2)

1−2y exp(x2+y2)

Implicit examples


ddx

(x + sin(x)) = ddx

(y − cos(y))

1 + cos(x) = dydx

+ sin(y) dydx

dydx

= 1+cos(x)1+sin(y)


dydx

= ddx

exp(x2 + y 2) = ddx

(ex2

ey2

)

= 2xex2

ey2

+ ex2

.2y dydxey

2

= 2(x + y dydx

) exp(x2 + y 2)

(1− 2y exp(x2 + y 2)) dydx

= 2x exp(x2 + y 2)

dydx

= 2x exp(x2+y2)

1−2y exp(x2+y2)

Implicit examples


ddx

(x + sin(x)) = ddx

(y − cos(y))

1 + cos(x) = dydx

+ sin(y) dydx

dydx

= 1+cos(x)1+sin(y)


dydx

= ddx

exp(x2 + y 2) = ddx

(ex2

ey2

)

= 2xex2

ey2

+ ex2

.2y dydxey

2

= 2(x + y dydx

) exp(x2 + y 2)

(1− 2y exp(x2 + y 2)) dydx

= 2x exp(x2 + y 2)

dydx

= 2x exp(x2+y2)

1−2y exp(x2+y2)

Implicit examples


ddx

(x + sin(x)) = ddx

(y − cos(y))

1 + cos(x) = dydx

+ sin(y) dydx

dydx

= 1+cos(x)1+sin(y)


dydx

= ddx

exp(x2 + y 2) = ddx

(ex2

ey2

)

= 2xex2

ey2

+ ex2

.2y dydxey

2

= 2(x + y dydx

) exp(x2 + y 2)

(1− 2y exp(x2 + y 2)) dydx

= 2x exp(x2 + y 2)

dydx

= 2x exp(x2+y2)

1−2y exp(x2+y2)

Parametric differentiation

I Suppose that x and y are both functions of another variable t. Then

dy

dx=

dy/dt

dx/dt

I Suppose that x = 1 + t2 and y = t + t3

(so t = y/x)

dy/dt = 1 + 3t2 dx/dt = 2t

dy

dx=

dy/dt

dx/dt=

1 + 3t2

2t=

1 + 3(y/x)2

2(y/x)=

x2 + 3y 2

2xy

I Suppose that x = t − sin(t) and y = 1− cos(t).

dy/dt = sin(t) dx/dt = 1− cos(t)

dy

dx=

dy/dt

dx/dt=

sin(t)

1− cos(t)=

√y(2− y)

y=

√2− y

y



dy

dx=

dy/dt

dx/dt


(so t = y/x)

dy/dt = 1 + 3t2 dx/dt = 2t

dy

dx=

dy/dt

dx/dt=

1 + 3t2

2t=

1 + 3(y/x)2

2(y/x)=

x2 + 3y 2

2xy



dy

dx=

dy/dt

dx/dt=

sin(t)

1− cos(t)=

√y(2− y)

y=

√2− y

y



dy

dx=

dy/dt

dx/dt


(so t = y/x)

dy/dt = 1 + 3t2 dx/dt = 2t

dy

dx=

dy/dt

dx/dt=

1 + 3t2

2t=

1 + 3(y/x)2

2(y/x)=

x2 + 3y 2

2xy



dy

dx=

dy/dt

dx/dt=

sin(t)

1− cos(t)=

√y(2− y)

y=

√2− y

y



dy

dx=

dy/dt

dx/dt


(so t = y/x)

dy/dt = 1 + 3t2 dx/dt = 2t

dy

dx=

dy/dt

dx/dt=

1 + 3t2

2t=

1 + 3(y/x)2

2(y/x)=

x2 + 3y 2

2xy



dy

dx=

dy/dt

dx/dt=

sin(t)

1− cos(t)=

√y(2− y)

y=

√2− y

y



dy

dx=

dy/dt

dx/dt


(so t = y/x)

dy/dt = 1 + 3t2 dx/dt = 2t

dy

dx=

dy/dt

dx/dt

=1 + 3t2

2t=

1 + 3(y/x)2

2(y/x)=

x2 + 3y 2

2xy



dy

dx=

dy/dt

dx/dt=

sin(t)

1− cos(t)=

√y(2− y)

y=

√2− y

y



dy

dx=

dy/dt

dx/dt


(so t = y/x)

dy/dt = 1 + 3t2 dx/dt = 2t

dy

dx=

dy/dt

dx/dt=

1 + 3t2

2t

=1 + 3(y/x)2

2(y/x)=

x2 + 3y 2

2xy



dy

dx=

dy/dt

dx/dt=

sin(t)

1− cos(t)=

√y(2− y)

y=

√2− y

y



dy

dx=

dy/dt

dx/dt

I Suppose that x = 1 + t2 and y = t + t3 (so t = y/x)

dy/dt = 1 + 3t2 dx/dt = 2t

dy

dx=

dy/dt

dx/dt=

1 + 3t2

2t=

1 + 3(y/x)2

2(y/x)

=x2 + 3y 2

2xy



dy

dx=

dy/dt

dx/dt=

sin(t)

1− cos(t)=

√y(2− y)

y=

√2− y

y



dy

dx=

dy/dt

dx/dt


dy/dt = 1 + 3t2 dx/dt = 2t

dy

dx=

dy/dt

dx/dt=

1 + 3t2

2t=

1 + 3(y/x)2

2(y/x)=

x2 + 3y 2

2xy



dy

dx=

dy/dt

dx/dt=

sin(t)

1− cos(t)=

√y(2− y)

y=

√2− y

y



dy

dx=

dy/dt

dx/dt


dy/dt = 1 + 3t2 dx/dt = 2t

dy

dx=

dy/dt

dx/dt=

1 + 3t2

2t=

1 + 3(y/x)2

2(y/x)=

x2 + 3y 2

2xy



dy

dx=

dy/dt

dx/dt=

sin(t)

1− cos(t)=

√y(2− y)

y=

√2− y

y



dy

dx=

dy/dt

dx/dt


dy/dt = 1 + 3t2 dx/dt = 2t

dy

dx=

dy/dt

dx/dt=

1 + 3t2

2t=

1 + 3(y/x)2

2(y/x)=

x2 + 3y 2

2xy



dy

dx=

dy/dt

dx/dt=

sin(t)

1− cos(t)=

√y(2− y)

y=

√2− y

y



dy

dx=

dy/dt

dx/dt


dy/dt = 1 + 3t2 dx/dt = 2t

dy

dx=

dy/dt

dx/dt=

1 + 3t2

2t=

1 + 3(y/x)2

2(y/x)=

x2 + 3y 2

2xy



dy

dx=

dy/dt

dx/dt=

sin(t)

1− cos(t)

=

√y(2− y)

y=

√2− y

y



dy

dx=

dy/dt

dx/dt


dy/dt = 1 + 3t2 dx/dt = 2t

dy

dx=

dy/dt

dx/dt=

1 + 3t2

2t=

1 + 3(y/x)2

2(y/x)=

x2 + 3y 2

2xy



dy

dx=

dy/dt

dx/dt=

sin(t)

1− cos(t)=

√y(2− y)

y

=

√2− y

y



dy

dx=

dy/dt

dx/dt


dy/dt = 1 + 3t2 dx/dt = 2t

dy

dx=

dy/dt

dx/dt=

1 + 3t2

2t=

1 + 3(y/x)2

2(y/x)=

x2 + 3y 2

2xy



dy

dx=

dy/dt

dx/dt=

sin(t)

1− cos(t)=

√y(2− y)

y=

√2− y

y

The circle

I Consider a point (x , y) on the unit circle, so x2 + y 2 = 1.

I Differentiate x2 + y 2 = 1;

2x + 2ydy

dx= 0;

dy

dx= −2x

2y= −x

y

I Parametrically: x = cos(t), y = sin(t).

dy

dx=

dy/dt

dx/dt=

cos(t)

− sin(t)= −x

y

I Directly: y = (1− x2)1/2

dy

dx=

1

2(1− x2)−1/2 d

dx(1− x2) =

1

2y−1.(−2x) = −x

y.

The circle



2x + 2ydy

dx= 0;

dy

dx= −2x

2y= −x

y


dy

dx=

dy/dt

dx/dt=

cos(t)

− sin(t)= −x

y


dy

dx=

1

2(1− x2)−1/2 d

dx(1− x2) =

1

2y−1.(−2x) = −x

y.

The circle



2x + 2ydy

dx= 0;

dy

dx= −2x

2y= −x

y


dy

dx=

dy/dt

dx/dt=

cos(t)

− sin(t)= −x

y


dy

dx=

1

2(1− x2)−1/2 d

dx(1− x2) =

1

2y−1.(−2x) = −x

y.

The circle


I Differentiate x2 + y 2 = 1; 2x + 2ydy

dx= 0;

dy

dx= −2x

2y= −x

y


dy

dx=

dy/dt

dx/dt=

cos(t)

− sin(t)= −x

y


dy

dx=

1

2(1− x2)−1/2 d

dx(1− x2) =

1

2y−1.(−2x) = −x

y.

The circle



dx= 0;

dy

dx= −2x

2y= −x

y


dy

dx=

dy/dt

dx/dt=

cos(t)

− sin(t)= −x

y


dy

dx=

1

2(1− x2)−1/2 d

dx(1− x2) =

1

2y−1.(−2x) = −x

y.

The circle



dx= 0;

dy

dx= −2x

2y= −x

y


dy

dx=

dy/dt

dx/dt=

cos(t)

− sin(t)= −x

y


dy

dx=

1

2(1− x2)−1/2 d

dx(1− x2) =

1

2y−1.(−2x) = −x

y.

The circle



dx= 0;

dy

dx= −2x

2y= −x

y


dy

dx=

dy/dt

dx/dt

=cos(t)

− sin(t)= −x

y


dy

dx=

1

2(1− x2)−1/2 d

dx(1− x2) =

1

2y−1.(−2x) = −x

y.

The circle



dx= 0;

dy

dx= −2x

2y= −x

y


dy

dx=

dy/dt

dx/dt=

cos(t)

− sin(t)

= −x

y


dy

dx=

1

2(1− x2)−1/2 d

dx(1− x2) =

1

2y−1.(−2x) = −x

y.

The circle



dx= 0;

dy

dx= −2x

2y= −x

y


dy

dx=

dy/dt

dx/dt=

cos(t)

− sin(t)= −x

y


dy

dx=

1

2(1− x2)−1/2 d

dx(1− x2) =

1

2y−1.(−2x) = −x

y.

The circle



dx= 0;

dy

dx= −2x

2y= −x

y


dy

dx=

dy/dt

dx/dt=

cos(t)

− sin(t)= −x

y


dy

dx=

1

2(1− x2)−1/2 d

dx(1− x2) =

1

2y−1.(−2x) = −x

y.

The circle



dx= 0;

dy

dx= −2x

2y= −x

y


dy

dx=

dy/dt

dx/dt=

cos(t)

− sin(t)= −x

y


dy

dx=

1

2(1− x2)−1/2 d

dx(1− x2)

=1

2y−1.(−2x) = −x

y.

The circle



dx= 0;

dy

dx= −2x

2y= −x

y


dy

dx=

dy/dt

dx/dt=

cos(t)

− sin(t)= −x

y


dy

dx=

1

2(1− x2)−1/2 d

dx(1− x2) =

1

2y−1.(−2x)

= −x

y.

The circle



dx= 0;

dy

dx= −2x

2y= −x

y


dy

dx=

dy/dt

dx/dt=

cos(t)

− sin(t)= −x

y


dy

dx=

1

2(1− x2)−1/2 d

dx(1− x2) =

1

2y−1.(−2x) = −x

y.

Integration


I The meaning of integration (take the sum of a large number of very smallcontributions, and pass to the limit)

I Integration as the reverse of differentiation

I Integrals of standard functions and classes of functions

I The method of undetermined coefficients

I Integration by parts

I Integration by substitution

Integration








Integration








Integration








Integration








Integration








Integration








Integration








Meaning

I To define∫ b

af (x) dx :

I Divide the interval [a, b] into many short intervals [x , x + h].I For each short interval [x , x + h], find f (x)h.I Add these terms together to get an approximation to

∫ ba f (x) dx .

I For the exact value of∫ ba f (x) dx , take the limit h −→ 0.

I In economics, government revenue depends on time, and total revenue inthe last decade is

∫ 2009

1999revenue(t) dt.

I If a particle moves with velocity v(t) > 0 at time t, then the total distance

moved between times a and b is∫ b

av(t) dt.

I A current flowing in a wire exerts a magnetic force on a moving electron.There is a formula for the force contributed by a short section of wire; toget the total force, we integrate.

Meaning

I To define∫ b

af (x) dx :


∫ ba f (x) dx .



∫ 2009

1999revenue(t) dt.



av(t) dt.


Meaning

I To define∫ b

af (x) dx :

I Divide the interval [a, b] into many short intervals [x , x + h].

I For each short interval [x , x + h], find f (x)h.I Add these terms together to get an approximation to

∫ ba f (x) dx .



∫ 2009

1999revenue(t) dt.



av(t) dt.


Meaning

I To define∫ b

af (x) dx :

I Divide the interval [a, b] into many short intervals [x , x + h].I For each short interval [x , x + h], find f (x)h.

I Add these terms together to get an approximation to∫ ba f (x) dx .



∫ 2009

1999revenue(t) dt.



av(t) dt.


Meaning

I To define∫ b

af (x) dx :


∫ ba f (x) dx .



∫ 2009

1999revenue(t) dt.



av(t) dt.


Meaning

I To define∫ b

af (x) dx :


∫ ba f (x) dx .



∫ 2009

1999revenue(t) dt.



av(t) dt.


Meaning

I To define∫ b

af (x) dx :


∫ ba f (x) dx .



∫ 2009

1999revenue(t) dt.



av(t) dt.


Meaning

I To define∫ b

af (x) dx :


∫ ba f (x) dx .



∫ 2009

1999revenue(t) dt.



av(t) dt.


Meaning

I To define∫ b

af (x) dx :


∫ ba f (x) dx .



∫ 2009

1999revenue(t) dt.



av(t) dt.


Areas

a b

y=f (x)

Consider the integral∫ b

af (x) dx .

Areas

a b

y=f (x)

f (x)

h

Area = f (x)h

For each short interval [x , x + h] ⊂ [a, b], we have a contribution f (x)h. This isthe area of the green rectangle.

Areas

a b

y=f (x)

This is the contribution from one short interval, but we need to add togetherthe contributions from many short intervals.

Areas

a b

y=f (x)

Here we have added in two more intervals

Areas

a b

y=f (x)

Here we have added in two more intervals – and two more

Areas

a b

y=f (x)

Here we have added in two more intervals – and two more – and two more

Areas

a b

y=f (x)

Now we have divided the whole interval [a, b] into subintervals of length h. Thesum of the terms f (x)h is the area of the green region.

Areas

a b

y=f (x)

This is not exactly the same as the area under the curve, because of the regionsmarked in blue and pink.

Areas

a b

y=f (x)

However, the error decreases if we make h smaller.

Areas

a b

y=f (x)

However, the error decreases if we make h smaller, and tends to zero in thelimit.

The Fundamental Theorem of Calculus

I An indefinite integral of f (x) is a function F (x) such that F ′(x) = f (x).I Examples:

I log(x) is an indefinite integral of 1/xI sin(x) is an indefinite integral of cos(x)I F (x) = x2 + 2x and G(x) = (x + 1)2 are indefinite integrals of 2x + 2

I The Fundamental Theorem of Calculus:

I For any number a, the function F (x) =∫ xa f (t) dt is an indefinite integral of

f (x).I If F (x) is any indefinite integral of f (x), then∫ b

a f (x) dx =[F (x)

]ba

= F (b)− F (a).

I The functions F (x) =∫ x

02t + 2 dt = x2 + 2x and

G(x) =∫ x

−12t + 2 dt = (x + 1)2 are both indefinite integrals of 2x + 2.

I∫ b

a1x

=[

log(x)]ba

= log(b)− log(a)


I An indefinite integral of f (x) is a function F (x) such that F ′(x) = f (x).

I Examples:





a f (x) dx =[F (x)

]ba

= F (b)− F (a).


02t + 2 dt = x2 + 2x and

G(x) =∫ x


I∫ b

a1x

=[

log(x)]ba

= log(b)− log(a)







a f (x) dx =[F (x)

]ba

= F (b)− F (a).


02t + 2 dt = x2 + 2x and

G(x) =∫ x


I∫ b

a1x

=[

log(x)]ba

= log(b)− log(a)



I log(x) is an indefinite integral of 1/x

I sin(x) is an indefinite integral of cos(x)I F (x) = x2 + 2x and G(x) = (x + 1)2 are indefinite integrals of 2x + 2




a f (x) dx =[F (x)

]ba

= F (b)− F (a).


02t + 2 dt = x2 + 2x and

G(x) =∫ x


I∫ b

a1x

=[

log(x)]ba

= log(b)− log(a)



I log(x) is an indefinite integral of 1/xI sin(x) is an indefinite integral of cos(x)

I F (x) = x2 + 2x and G(x) = (x + 1)2 are indefinite integrals of 2x + 2




a f (x) dx =[F (x)

]ba

= F (b)− F (a).


02t + 2 dt = x2 + 2x and

G(x) =∫ x


I∫ b

a1x

=[

log(x)]ba

= log(b)− log(a)







a f (x) dx =[F (x)

]ba

= F (b)− F (a).


02t + 2 dt = x2 + 2x and

G(x) =∫ x


I∫ b

a1x

=[

log(x)]ba

= log(b)− log(a)







a f (x) dx =[F (x)

]ba

= F (b)− F (a).


02t + 2 dt = x2 + 2x and

G(x) =∫ x


I∫ b

a1x

=[

log(x)]ba

= log(b)− log(a)




I The Fundamental Theorem of Calculus:I For any number a, the function F (x) =

∫ xa f (t) dt is an indefinite integral of

f (x).

I If F (x) is any indefinite integral of f (x), then∫ ba f (x) dx =

[F (x)

]ba

= F (b)− F (a).


02t + 2 dt = x2 + 2x and

G(x) =∫ x


I∫ b

a1x

=[

log(x)]ba

= log(b)− log(a)







a f (x) dx =[F (x)

]ba

= F (b)− F (a).


02t + 2 dt = x2 + 2x and

G(x) =∫ x


I∫ b

a1x

=[

log(x)]ba

= log(b)− log(a)







a f (x) dx =[F (x)

]ba

= F (b)− F (a).


02t + 2 dt = x2 + 2x and

G(x) =∫ x


I∫ b

a1x

=[

log(x)]ba

= log(b)− log(a)







a f (x) dx =[F (x)

]ba

= F (b)− F (a).


02t + 2 dt = x2 + 2x and

G(x) =∫ x


I∫ b

a1x

=[

log(x)]ba

= log(b)− log(a)

Proof of the Fundamental Theorem

Area = F (x)

y=f (x)

a x

Choose a number a, and define F (x) =∫ x

af (t) dt. We must show that

F ′(x) = f (x).

We now change x to x + h. The increase in F (x) isF (x + h)− F (x), which is the area of the thin strip as shown. This isapproximately the same as f (x)h.


Area = F (x + h)

x+h

y=f (x)

a x

We now change x to x + h.

The increase in F (x) is F (x + h)− F (x), which isthe area of the thin strip as shown. This is approximately the same as f (x)h.


x+h

Area = F (x + h)− F (x)

' f (x)h

y=f (x)

a x

We now change x to x + h. The increase in F (x) is F (x + h)− F (x), which isthe area of the thin strip as shown.

This is approximately the same as f (x)h.


x+h

Area = F (x + h)− F (x)' f (x)h

f (x)

h

y=f (x)

a x

We now change x to x + h. The increase in F (x) is F (x + h)− F (x), which isthe area of the thin strip as shown. This is approximately the same as f (x)h.


Area = F (x + h)− F (x)' f (x)h

x+h

f (x)

h

y=f (x)

a x



Area = F (x + h)− F (x)' f (x)h

x+h

f (x)

h

y=f (x)

a x


F ′(x) ' (F (x + h)− F (x))/h ' f (x).


Area = F (x + h)− F (x)' f (x)h

x+h

f (x)

h

y=f (x)

a x


F ′(x) = limh−→0

(F (x + h)− F (x))/h = f (x).

Constants

I Is it

∫x2 dx = x3/3 or

∫x2 dx = x3/3+c?

I Either is acceptable in the exam.Neither one is strictly logically satisfactory.

I x3/3 is an indefinite integral of x2.

I Every indefinite integral of x2 has the form x3/3 + c for some c.

I If you just want to calculate∫ b

af (x) dx , it does not matter which

indefinite integral you use. Any two choices will give the same answer.

I In solving differential equations, it often does matter which indefiniteintegral you use. You must therefore include a ’+c’ term, and do someextra work to see what c should be.

I Maple’s int() command will never give you a ’+c’ term.If you need one, you must insert it yourself.

Constants

I Is it

∫x2 dx = x3/3 or

∫x2 dx = x3/3+c?









Constants

I Is it

∫x2 dx = x3/3 or

∫x2 dx = x3/3+c?









Constants

I Is it

∫x2 dx = x3/3 or

∫x2 dx = x3/3+c?









Constants

I Is it

∫x2 dx = x3/3 or

∫x2 dx = x3/3+c?









Constants

I Is it

∫x2 dx = x3/3 or

∫x2 dx = x3/3+c?









Constants

I Is it

∫x2 dx = x3/3 or

∫x2 dx = x3/3+c?









Constants

I Is it

∫x2 dx = x3/3 or

∫x2 dx = x3/3+c?









Checking and Guessing

IIntegrals can easily be checked by differentiating

I∫

sin(x)2 dx 6= sin(x)3/3, because

d

dx

(sin(x)3/3

)= 3 sin(x)2 cos(x)/3 = sin(x)2 cos(x)6= sin(x)2.

I∫ cos(x)

x− sin(x)

x2 dx = sin(x)x

, because

d

dx

(sin(x)

x

)=

sin′(x).x − sin(x).1

x2=

cos(x)

x− sin(x)

x2.

I∫

2x ex2

dx = ex2

, because ddxex

2

= 2x ex2

.

I∫

3x2+2x+1x3+x2+x+1

dx = log(x3 + x2 + x + 1)

, because

d

dxlog(x3 + x2 + x + 1) =

ddx

(x3 + x2 + x + 1)

x3 + x2 + x + 1=

3x2 + 2x + 1

x3 + x2 + x + 1.



I∫


d

dx

(sin(x)3/3

)= 3 sin(x)2 cos(x)/3 = sin(x)2 cos(x)6= sin(x)2.

I∫ cos(x)

x− sin(x)

x2 dx = sin(x)x

, because

d

dx

(sin(x)

x

)=


x2=

cos(x)

x− sin(x)

x2.

I∫

2x ex2

dx = ex2

, because ddxex

2

= 2x ex2

.

I∫

3x2+2x+1x3+x2+x+1

dx = log(x3 + x2 + x + 1)

, because

d

dxlog(x3 + x2 + x + 1) =

ddx

(x3 + x2 + x + 1)

x3 + x2 + x + 1=

3x2 + 2x + 1

x3 + x2 + x + 1.



I∫

sin(x)2 dx = sin(x)3/3?

∫sin(x)2 dx 6= sin(x)3/3, because

d

dx

(sin(x)3/3

)= 3 sin(x)2 cos(x)/3 = sin(x)2 cos(x) 6= sin(x)2.

I∫ cos(x)

x− sin(x)

x2 dx = sin(x)x

, because

d

dx

(sin(x)

x

)=


x2=

cos(x)

x− sin(x)

x2.

I∫

2x ex2

dx = ex2

, because ddxex

2

= 2x ex2

.

I∫

3x2+2x+1x3+x2+x+1

dx = log(x3 + x2 + x + 1)

, because

d

dxlog(x3 + x2 + x + 1) =

ddx

(x3 + x2 + x + 1)

x3 + x2 + x + 1=

3x2 + 2x + 1

x3 + x2 + x + 1.



I∫


d

dx

(sin(x)3/3


I∫ cos(x)

x− sin(x)

x2 dx = sin(x)x

, because

d

dx

(sin(x)

x

)=


x2=

cos(x)

x− sin(x)

x2.

I∫

2x ex2

dx = ex2

, because ddxex

2

= 2x ex2

.

I∫

3x2+2x+1x3+x2+x+1

dx = log(x3 + x2 + x + 1)

, because

d

dxlog(x3 + x2 + x + 1) =

ddx

(x3 + x2 + x + 1)

x3 + x2 + x + 1=

3x2 + 2x + 1

x3 + x2 + x + 1.



I∫


d

dx

(sin(x)3/3


I∫ cos(x)

x− sin(x)

x2 dx = sin(x)x

?

∫ cos(x)x− sin(x)

x2 dx = sin(x)x

, because

d

dx

(sin(x)

x

)=


x2=

cos(x)

x− sin(x)

x2.

I∫

2x ex2

dx = ex2

, because ddxex

2

= 2x ex2

.

I∫

3x2+2x+1x3+x2+x+1

dx = log(x3 + x2 + x + 1)

, because

d

dxlog(x3 + x2 + x + 1) =

ddx

(x3 + x2 + x + 1)

x3 + x2 + x + 1=

3x2 + 2x + 1

x3 + x2 + x + 1.



I∫


d

dx

(sin(x)3/3


I∫ cos(x)

x− sin(x)

x2 dx = sin(x)x

, because

d

dx

(sin(x)

x

)=


x2=

cos(x)

x− sin(x)

x2.

I∫

2x ex2

dx = ex2

, because ddxex

2

= 2x ex2

.

I∫

3x2+2x+1x3+x2+x+1

dx = log(x3 + x2 + x + 1)

, because

d

dxlog(x3 + x2 + x + 1) =

ddx

(x3 + x2 + x + 1)

x3 + x2 + x + 1=

3x2 + 2x + 1

x3 + x2 + x + 1.



I∫


d

dx

(sin(x)3/3


I∫ cos(x)

x− sin(x)

x2 dx = sin(x)x

, because

d

dx

(sin(x)

x

)=


x2=

cos(x)

x− sin(x)

x2.

I∫

2x ex2

dx = ex2

, because ddxex

2

= 2x ex2

.

I∫

3x2+2x+1x3+x2+x+1

dx = log(x3 + x2 + x + 1)

, because

d

dxlog(x3 + x2 + x + 1) =

ddx

(x3 + x2 + x + 1)

x3 + x2 + x + 1=

3x2 + 2x + 1

x3 + x2 + x + 1.



I∫


d

dx

(sin(x)3/3


I∫ cos(x)

x− sin(x)

x2 dx = sin(x)x

, because

d

dx

(sin(x)

x

)=


x2=

cos(x)

x− sin(x)

x2.

I∫

2x ex2

dx = ex2

, because ddxex

2

= 2x ex2

.

I∫

3x2+2x+1x3+x2+x+1

dx = log(x3 + x2 + x + 1)

, because

d

dxlog(x3 + x2 + x + 1) =

ddx

(x3 + x2 + x + 1)

x3 + x2 + x + 1=

3x2 + 2x + 1

x3 + x2 + x + 1.



I∫


d

dx

(sin(x)3/3


I∫ cos(x)

x− sin(x)

x2 dx = sin(x)x

, because

d

dx

(sin(x)

x

)=


x2=

cos(x)

x− sin(x)

x2.

I∫

2x ex2

dx = ex2

, because ddxex

2

= 2x ex2

.

I∫

3x2+2x+1x3+x2+x+1

dx = log(x3 + x2 + x + 1)

, because

d

dxlog(x3 + x2 + x + 1) =

ddx

(x3 + x2 + x + 1)

x3 + x2 + x + 1=

3x2 + 2x + 1

x3 + x2 + x + 1.



I∫


d

dx

(sin(x)3/3


I∫ cos(x)

x− sin(x)

x2 dx = sin(x)x

, because

d

dx

(sin(x)

x

)=


x2=

cos(x)

x− sin(x)

x2.

I∫

2x ex2

dx = ex2

, because ddxex

2

= 2x ex2

.

I∫

3x2+2x+1x3+x2+x+1

dx = log(x3 + x2 + x + 1), because

d

dxlog(x3 + x2 + x + 1) =

ddx

(x3 + x2 + x + 1)

x3 + x2 + x + 1=

3x2 + 2x + 1

x3 + x2 + x + 1.

Undetermined coefficients

I Suppose we know that for some constants a, . . . , d∫log(x)3 dx = (a log(x)3 + b log(x)2 + c log(x) + d)x

(How could we know this? — see later)

I Problem: find a, b, c and d .

I . log(x)3 = ddx

((a log(x)3 + b log(x)2 + c log(x) + d)x

)

= (3a log(x)2x−1 + 2b log(x)x−1 + cx−1)x +

(a log(x)3 + b log(x)2 + c log(x) + d).1

= a log(x)3 + (b + 3a) log(x)2 + (c + 2b) log(x) + (d + c)

I So a = 1, b + 3a = 0, c + 2b = 0 and d + c = 0 (compare coefficients)

I So a = 1, b = −3, c = 6 and d = −6∫log(x)3 dx = (log(x)3 − 3 log(x)2 + 6 log(x)− 6)x .



(How could we know this? — see later)I Problem: find a, b, c and d .

I . log(x)3 = ddx


)








(How could we know this? — see later)

I Problem: find a, b, c and d .

I . log(x)3 = ddx


)









I . log(x)3 = ddx


)









I . log(x)3 = ddx


)









I . log(x)3 = ddx


)= (3a log(x)2x−1 + 2b log(x)x−1 + cx−1)x +








I . log(x)3 = ddx










I . log(x)3 = ddx










I . log(x)3 = ddx






I So a = 1, b = −3, c = 6 and d = −6

∫log(x)3 dx = (log(x)3 − 3 log(x)2 + 6 log(x)− 6)x .




I . log(x)3 = ddx







Standard integrals

= == == == == == == =∫

xn dx = xn+1/(n + 1) (n 6= −1)∫ax dx = ax/ log(a)∫

log(x) dx = x log(x)− x∫tan(x) dx = − log(cos(x))∫sin(x)2 dx = (2x − sin(2x))/4∫cos(x)2 dx = (2x + sin(2x))/4

Standard integrals




tanh′(x) = sech(x)2 arctanh′(x) = (1− x2)−1



tan′(x) = sec(x)2 arctan′(x) = (1 + x2)−1

∫xn dx = xn+1/(n + 1) (n 6= −1)∫ax dx = ax/ log(a)∫


Standard integrals

∫exp(x) dx = exp(x)

∫1/x dx = log(x)∫

cosh(x) dx = sinh(x)∫

(1 + x2)−1/2 dx = arcsinh(x)∫sinh(x) dx = cosh(x)

∫(x2 − 1)−1/2 dx = arccosh(x)∫

sech(x)2 dx = tanh(x)∫

(1− x2)−1 dx = arctanh(x)∫cos(x) dx = sin(x)

∫(1− x2)−1/2 dx = arcsin(x)∫

sin(x) dx = − cos(x)∫

(1− x2)−1/2 dx = − arccos(x)∫sec(x)2 dx = tan(x)

∫(1 + x2)−1 dx = arctan(x)

∫xn dx = xn+1/(n + 1) (n 6= −1)∫ax dx = ax/ log(a)∫


Standard integrals





∫(x2 − 1)−1/2 dx = arccosh(x)∫



∫(1− x2)−1/2 dx = arcsin(x)∫



∫(1 + x2)−1 dx = arctan(x)∫

xn dx = xn+1/(n + 1) (n 6= −1)

∫ax dx = ax/ log(a)∫


Standard integrals





∫(x2 − 1)−1/2 dx = arccosh(x)∫



∫(1− x2)−1/2 dx = arcsin(x)∫



∫(1 + x2)−1 dx = arctan(x)∫

xn dx = xn+1/(n + 1) (n 6= −1)∫ax dx = ax/ log(a)

∫log(x) dx = x log(x)− x∫tan(x) dx = − log(cos(x))∫sin(x)2 dx = (2x − sin(2x))/4∫cos(x)2 dx = (2x + sin(2x))/4

Standard integrals





∫(x2 − 1)−1/2 dx = arccosh(x)∫



∫(1− x2)−1/2 dx = arcsin(x)∫



∫(1 + x2)−1 dx = arctan(x)∫


log(x) dx = x log(x)− x

∫tan(x) dx = − log(cos(x))∫sin(x)2 dx = (2x − sin(2x))/4∫cos(x)2 dx = (2x + sin(2x))/4

Standard integrals





∫(x2 − 1)−1/2 dx = arccosh(x)∫



∫(1− x2)−1/2 dx = arcsin(x)∫



∫(1 + x2)−1 dx = arctan(x)∫


log(x) dx = x log(x)− x∫tan(x) dx = − log(cos(x))

∫sin(x)2 dx = (2x − sin(2x))/4∫cos(x)2 dx = (2x + sin(2x))/4

Standard integrals





∫(x2 − 1)−1/2 dx = arccosh(x)∫



∫(1− x2)−1/2 dx = arcsin(x)∫



∫(1 + x2)−1 dx = arctan(x)∫


log(x) dx = x log(x)− x∫tan(x) dx = − log(cos(x))∫sin(x)2 dx = (2x − sin(2x))/4

∫cos(x)2 dx = (2x + sin(2x))/4

Standard integrals





∫(x2 − 1)−1/2 dx = arccosh(x)∫



∫(1− x2)−1/2 dx = arcsin(x)∫



∫(1 + x2)−1 dx = arctan(x)∫



Rational functions

I A rational function of x is a function defined using only constants,addition, multiplication, division and integer powers.

I No roots, fractional powers, logs, exponentials, trigonometric functionsand so on can occur in a rational function.

I Examples: 1 + x + x2

1 − x + x2

1

x+

π

x − 1+

π2

x − 2x2 + x + 1 + x−1 + x−2

I Non-Examples: e−x sin(x)

√1 − x2

log(x)

1 + x

arctan(x)

2π.

I If f (x) is a rational function, then∫f (x) dx is a sum of terms of the

following types:I Rational functionsI Terms of the form ln(|x − u|)I Terms of the form ln(x2 + vx + w)I Terms of the form arctan(ux + v).

I∫

4x3 + 8

x6 − x2dx =

8

x+ 3 ln(|x − 1|)− ln(|x + 1|)− ln(x2 + 1) + 4 arctan(x)

Rational functions




1 − x + x2

1

x+

π

x − 1+

π2

x − 2x2 + x + 1 + x−1 + x−2


√1 − x2

log(x)

1 + x

arctan(x)

2π.



I∫

4x3 + 8

x6 − x2dx =

8


Rational functions




1 − x + x2

1

x+

π

x − 1+

π2

x − 2x2 + x + 1 + x−1 + x−2


√1 − x2

log(x)

1 + x

arctan(x)

2π.



I∫

4x3 + 8

x6 − x2dx =

8


Rational functions




1 − x + x2

1

x+

π

x − 1+

π2

x − 2x2 + x + 1 + x−1 + x−2


√1 − x2

log(x)

1 + x

arctan(x)

2π.



I∫

4x3 + 8

x6 − x2dx =

8


Rational functions




1 − x + x2

1

x+

π

x − 1+

π2

x − 2x2 + x + 1 + x−1 + x−2


√1 − x2

log(x)

1 + x

arctan(x)

2π.



I∫

4x3 + 8

x6 − x2dx =

8


Rational functions




1 − x + x2

1

x+

π

x − 1+

π2

x − 2x2 + x + 1 + x−1 + x−2


√1 − x2

log(x)

1 + x

arctan(x)

2π.



I∫

4x3 + 8

x6 − x2dx =

8


Rational functions




1 − x + x2

1

x+

π

x − 1+

π2

x − 2x2 + x + 1 + x−1 + x−2


√1 − x2

log(x)

1 + x

arctan(x)

2π.



I∫

4x3 + 8

x6 − x2dx =

8


Rational function examples

I∫

x2 + 1

x2 − 1dx = x + ln(|x − 1|) + ln(|x + 1|)

I∫ (

x + 1

x − 1

)3

dx = 1 +6

x − 1+

12

(x − 1)2+

8

(x − 1)3

I∫

2x + 2

x2 + 1dx = ln(x2 + 1) + 2 arctan(x)

I∫

1

x−1 + 1 + xdx =

1

2ln(1 + x + x2)− 1√

3arctan

(1 + 2x√

3

)I∫

4

1− x4dx = ln(|x + 1|)− ln(|x − 1|) + 2 arctan(x)

d

dxln(|x − u|) =

1

x − u

d

dxln(x2 + ux + v) =

2x + u

x2 + ux + v

d

dxarctan(ux + v) =

u

1 + (ux + v)2=

u

u2x2 + 2uvx + (v 2 + 1)


I∫

x2 + 1

x2 − 1dx = x + ln(|x − 1|) + ln(|x + 1|)

I∫ (

x + 1

x − 1

)3

dx = 1 +6

x − 1+

12

(x − 1)2+

8

(x − 1)3

I∫

2x + 2

x2 + 1dx = ln(x2 + 1) + 2 arctan(x)

I∫

1

x−1 + 1 + xdx =

1

2ln(1 + x + x2)− 1√

3arctan

(1 + 2x√

3

)I∫

4


d

dxln(|x − u|) =

1

x − u

d

dxln(x2 + ux + v) =

2x + u

x2 + ux + v

d

dxarctan(ux + v) =

u

1 + (ux + v)2=

u

u2x2 + 2uvx + (v 2 + 1)


I∫

x2 + 1

x2 − 1dx = x + ln(|x − 1|) + ln(|x + 1|)

I∫ (

x + 1

x − 1

)3

dx = 1 +6

x − 1+

12

(x − 1)2+

8

(x − 1)3

I∫

2x + 2

x2 + 1dx = ln(x2 + 1) + 2 arctan(x)

I∫

1

x−1 + 1 + xdx =

1

2ln(1 + x + x2)− 1√

3arctan

(1 + 2x√

3

)I∫

4


d

dxln(|x − u|) =

1

x − u

d

dxln(x2 + ux + v) =

2x + u

x2 + ux + v

d

dxarctan(ux + v) =

u

1 + (ux + v)2=

u

u2x2 + 2uvx + (v 2 + 1)

Trigonometric polynomials

∫sin(nx) dx = − cos(nx)/n

∫cos(nx) dx = sin(nx)/n

cos(2x) = cos(x)2 − sin(x)2 = 2 cos(x)2 − 1 = 1− 2 sin(x)2

sin(x)2 = 1/2− cos(2x)/2∫sin(x)2 dx = x/2− sin(2x)/4∫cos(x)2 dx = x/2 + sin(2x)/4

sin(x)3 = 3 sin(x)/4− sin(3x)/4∫sin(x)3 dx = −3 cos(x)/4 + cos(3x)/12

sin(x) sin(2x) sin(4x) = − sin(x)/4 + sin(3x)/4 + sin(5x)/4− sin(7x)/4∫sin(x) sin(2x) sin(4x) dx = cos(x)/4− cos(3x)/12− cos(5x)/20 + cos(7x)/28

sin(x)4 + cos(x)4 = 3/4 + cos(4x)/4∫sin(x)4 + cos(x)4 dx = 3x/4 + sin(4x)/16





















sin(x)2 = 1/2− cos(2x)/2

∫sin(x)2 dx = x/2− sin(2x)/4∫cos(x)2 dx = x/2 + sin(2x)/4








sin(x)2 = 1/2− cos(2x)/2∫sin(x)2 dx = x/2− sin(2x)/4

∫cos(x)2 dx = x/2 + sin(2x)/4

















sin(x)3 = 3 sin(x)/4− sin(3x)/4

∫sin(x)3 dx = −3 cos(x)/4 + cos(3x)/12

















sin(x) sin(2x) sin(4x) = − sin(x)/4 + sin(3x)/4 + sin(5x)/4− sin(7x)/4

∫sin(x) sin(2x) sin(4x) dx = cos(x)/4− cos(3x)/12− cos(5x)/20 + cos(7x)/28

















sin(x)4 + cos(x)4 = 3/4 + cos(4x)/4

∫sin(x)4 + cos(x)4 dx = 3x/4 + sin(4x)/16









Affine substitution

If

∫f (x) dx = g(x) and a, b are constant, then∫

f (ax + b) dx = g(ax + b)/a

∫cos(x) dx = sin(x)

∫cos(2x + 3) dx = sin(2x + 3)/2∫

ex dx = ex∫

e−2x+7 dx = e−2x+7/(−2)∫tan(x) dx = − ln(cos(x))

∫tan(πx) dx = − ln(cos(πx))/π

Affine substitution

If




∫cos(2x + 3) dx = sin(2x + 3)/2∫

ex dx = ex∫



Affine substitution

If




∫cos(2x + 3) dx = sin(2x + 3)/2∫

ex dx = ex∫



Affine substitution

If




∫cos(2x + 3) dx = sin(2x + 3)/2

∫ex dx = ex

∫e−2x+7 dx = e−2x+7/(−2)∫

tan(x) dx = − ln(cos(x))


Affine substitution

If




∫cos(2x + 3) dx = sin(2x + 3)/2∫

ex dx = ex

∫e−2x+7 dx = e−2x+7/(−2)∫

tan(x) dx = − ln(cos(x))


Affine substitution

If




∫cos(2x + 3) dx = sin(2x + 3)/2∫

ex dx = ex∫

e−2x+7 dx = e−2x+7/(−2)

∫tan(x) dx = − ln(cos(x))


Affine substitution

If




∫cos(2x + 3) dx = sin(2x + 3)/2∫

ex dx = ex∫



Affine substitution

If




∫cos(2x + 3) dx = sin(2x + 3)/2∫

ex dx = ex∫



Exponential oscillations

I An exponential oscillation is a function of the form

f (x) = eλx(a cos(ωx) + b sin(ωx)),

where a, b, λ and ω are constants.

I The growth rate is λ, and the angular frequency is ω.

y=e−x sin(20πx)

(λ=−1,ω=20π,a=0,b=1)

I Special cases:

f (x) = eλx sin(ωx) (a = 0, b = 1)

f (x) = a cos(ωx) + b sin(ωx) (λ = 0)

f (x) = aeλx (ω = 0).






y=e−x sin(20πx)

(λ=−1,ω=20π,a=0,b=1)

I Special cases:

f (x) = eλx sin(ωx) (a = 0, b = 1)


f (x) = aeλx (ω = 0).






y=e−x sin(20πx)

(λ=−1,ω=20π,a=0,b=1)

I Special cases:

f (x) = eλx sin(ωx) (a = 0, b = 1)


f (x) = aeλx (ω = 0).






y=e−x sin(20πx)

(λ=−1,ω=20π,a=0,b=1)

I Special cases:

f (x) = eλx sin(ωx) (a = 0, b = 1)


f (x) = aeλx (ω = 0).






y=e−x sin(20πx)

(λ=−1,ω=20π,a=0,b=1)

I Special cases:

f (x) = eλx sin(ωx) (a = 0, b = 1)


f (x) = aeλx (ω = 0).

Integrating exponential oscillations

The integral of an EO is another EO with the same growth rate and angularfrequency.∫

eλx(a cos(ωx) + b sin(ωx)) dx = eλx(A cos(ωx) + B sin(ωx))

A =aλ− bω

λ2 + ω2B =

aω + bλ

λ2 + ω2.

I Example: find

I λ = −2, ω = 4, a = 5, b = −3

I A =aλ− bω

λ2 + ω2=

5.(−2)− (−3).4

(−2)2 + 42= 1/10

I B =aω + bλ

λ2 + ω2=

5.4 + (−3)(−2)

(−2)2 + 42= 13/10∫

e−2x(5 cos(4x)− 3 sin(4x)) dx = e−2x(cos(4x) + 13 sin(4x))/10


The integral of an EO is another EO with the same growth rate and angularfrequency.

∫eλx(a cos(ωx) + b sin(ωx)) dx = eλx(A cos(ωx) + B sin(ωx))

A =aλ− bω

λ2 + ω2B =

aω + bλ

λ2 + ω2.

I Example: find

I λ = −2, ω = 4, a = 5, b = −3

I A =aλ− bω

λ2 + ω2=

5.(−2)− (−3).4

(−2)2 + 42= 1/10

I B =aω + bλ

λ2 + ω2=

5.4 + (−3)(−2)

(−2)2 + 42= 13/10∫





A =aλ− bω

λ2 + ω2B =

aω + bλ

λ2 + ω2.

I Example: find

I λ = −2, ω = 4, a = 5, b = −3

I A =aλ− bω

λ2 + ω2=

5.(−2)− (−3).4

(−2)2 + 42= 1/10

I B =aω + bλ

λ2 + ω2=

5.4 + (−3)(−2)

(−2)2 + 42= 13/10∫





A =aλ− bω

λ2 + ω2B =

aω + bλ

λ2 + ω2.

I Example: find

I λ = −2, ω = 4, a = 5, b = −3

I A =aλ− bω

λ2 + ω2=

5.(−2)− (−3).4

(−2)2 + 42= 1/10

I B =aω + bλ

λ2 + ω2=

5.4 + (−3)(−2)

(−2)2 + 42= 13/10∫





A =aλ− bω

λ2 + ω2B =

aω + bλ

λ2 + ω2.

I Example: find∫e−2x(5 cos(4x)−3 sin(4x)) dx

I λ = −2, ω = 4, a = 5, b = −3

I A =aλ− bω

λ2 + ω2=

5.(−2)− (−3).4

(−2)2 + 42= 1/10

I B =aω + bλ

λ2 + ω2=

5.4 + (−3)(−2)

(−2)2 + 42= 13/10∫





A =aλ− bω

λ2 + ω2B =

aω + bλ

λ2 + ω2.


I λ = −2, ω = 4, a = 5, b = −3

I A =aλ− bω

λ2 + ω2=

5.(−2)− (−3).4

(−2)2 + 42= 1/10

I B =aω + bλ

λ2 + ω2=

5.4 + (−3)(−2)

(−2)2 + 42= 13/10∫





A =aλ− bω

λ2 + ω2B =

aω + bλ

λ2 + ω2.


I λ = −2, ω = 4, a = 5, b = −3

I A =aλ− bω

λ2 + ω2=

5.(−2)− (−3).4

(−2)2 + 42= 1/10

I B =aω + bλ

λ2 + ω2=

5.4 + (−3)(−2)

(−2)2 + 42= 13/10∫





A =aλ− bω

λ2 + ω2B =

aω + bλ

λ2 + ω2.


I λ = −2, ω = 4, a = 5, b = −3

I A =aλ− bω

λ2 + ω2=

5.(−2)− (−3).4

(−2)2 + 42= 1/10

I B =aω + bλ

λ2 + ω2=

5.4 + (−3)(−2)

(−2)2 + 42= 13/10

∫e−2x(5 cos(4x)− 3 sin(4x)) dx = e−2x(cos(4x) + 13 sin(4x))/10




A =aλ− bω

λ2 + ω2B =

aω + bλ

λ2 + ω2.


I λ = −2, ω = 4, a = 5, b = −3

I A =aλ− bω

λ2 + ω2=

5.(−2)− (−3).4

(−2)2 + 42= 1/10

I B =aω + bλ

λ2 + ω2=

5.4 + (−3)(−2)

(−2)2 + 42= 13/10∫



Alternatively:∫e−2x(5 cos(4x)− 3 sin(4x)) dx = e−2x(A cos(4x) + B sin(4x)) for some A,B

e−2x(5 cos(4x)− 3 sin(4x)) =d

dx

(e−2x(A cos(4x) + B sin(4x))

)= − 2e−2x(A cos(4x) + B sin(4x)) +

e−2x(−4A sin(4x) + 4B cos(4x))

= e−2x((4B − 2A) cos(4x)− (2B + 4A) sin(4x))

By comparing coefficients, we must have 4B − 2A = 5 and 2B + 4A = 3.These equations can be solved to give A = 1/10 and B = 13/10. Thus∫

e−2x(5 cos(4x)− 3 sin(4x)) dx = e−2x(cos(4x) + 13 sin(4x))/10.



e−2x(5 cos(4x)− 3 sin(4x)) =d

dx


)= − 2e−2x(A cos(4x) + B sin(4x)) +


= e−2x((4B − 2A) cos(4x)− (2B + 4A) sin(4x))





e−2x(5 cos(4x)− 3 sin(4x)) =d

dx


)

= − 2e−2x(A cos(4x) + B sin(4x)) +


= e−2x((4B − 2A) cos(4x)− (2B + 4A) sin(4x))





e−2x(5 cos(4x)− 3 sin(4x)) =d

dx


)= − 2e−2x(A cos(4x) + B sin(4x)) +


= e−2x((4B − 2A) cos(4x)− (2B + 4A) sin(4x))





e−2x(5 cos(4x)− 3 sin(4x)) =d

dx


)= − 2e−2x(A cos(4x) + B sin(4x)) +


= e−2x((4B − 2A) cos(4x)− (2B + 4A) sin(4x))





e−2x(5 cos(4x)− 3 sin(4x)) =d

dx


)= − 2e−2x(A cos(4x) + B sin(4x)) +


= e−2x((4B − 2A) cos(4x)− (2B + 4A) sin(4x))

By comparing coefficients, we must have 4B − 2A = 5 and 2B + 4A = 3.

These equations can be solved to give A = 1/10 and B = 13/10. Thus∫e−2x(5 cos(4x)− 3 sin(4x)) dx = e−2x(cos(4x) + 13 sin(4x))/10.



e−2x(5 cos(4x)− 3 sin(4x)) =d

dx


)= − 2e−2x(A cos(4x) + B sin(4x)) +


= e−2x((4B − 2A) cos(4x)− (2B + 4A) sin(4x))

By comparing coefficients, we must have 4B − 2A = 5 and 2B + 4A = 3.These equations can be solved to give A = 1/10 and B = 13/10.

Thus∫e−2x(5 cos(4x)− 3 sin(4x)) dx = e−2x(cos(4x) + 13 sin(4x))/10.



e−2x(5 cos(4x)− 3 sin(4x)) =d

dx


)= − 2e−2x(A cos(4x) + B sin(4x)) +


= e−2x((4B − 2A) cos(4x)− (2B + 4A) sin(4x))



Polynomial exponential oscillations

I A polynomial exponential oscillation is a function of the form

f (x) = eλx(a(x) cos(ωx) + b(x) sin(ωx)),

where a(x) and b(x) are polynomials.

I λ is the growth rate and ω is the angular frequency. The degree is thehighest power of x that occurs in a(x) or in b(x).

I The function f (x) = e−2x((1 + x5) cos(4x) + x3 sin(4x))is a PEO of growth rate −2, frequency 4 and degree 5.

I The function f (x) = e4x((1 + x3 + x6) sin(3x))is a PEO of growth rate 4, frequency 3 and degree 6.

I Fact: The integral of any PEO is another PEO with the same growthrate, frequency and degree.









































Integrating PEO’s — I

I∫xe−x sin(x) dx is a PEO of degree 1, growth −1, frequency 1.

I∫xe−x sin(x) dx = (Ax + B)e−x cos(x) + (Cx + D)e−x sin(x)

for some A, B, C , D.

I .

xe−x sin(x) = ddx

((Ax + B)e−x cos(x) + (Cx + D)e−x sin(x)

)= Ae−x cos(x)− (Ax + B)e−x cos(x)− (Ax + B)e−x sin(x) +

Ce−x sin(x)− (Cx + D)e−x sin(x) + (Cx + D)e−x cos(x)

= (−A + C)xe−x cos(x) + (A− B + D)e−x cos(x) +

(−A− C)xe−x sin(x) + (−B + C − D)e−x sin(x).

I −A + C = 0, A− B + D = 0, −A− C = 1, −B + C − D = 0.

I So A = −1/2, B = −1/2, C = −1/2, D = 0

I∫xe−x sin(x) dx = −((x + 1)e−x cos(x) + xe−x sin(x))/2.





I .

xe−x sin(x) = ddx






I −A + C = 0, A− B + D = 0, −A− C = 1, −B + C − D = 0.

I So A = −1/2, B = −1/2, C = −1/2, D = 0






I .

xe−x sin(x) = ddx






I −A + C = 0, A− B + D = 0, −A− C = 1, −B + C − D = 0.

I So A = −1/2, B = −1/2, C = −1/2, D = 0






I .xe−x sin(x) = d

dx


)

= Ae−x cos(x)− (Ax + B)e−x cos(x)− (Ax + B)e−x sin(x) +




I −A + C = 0, A− B + D = 0, −A− C = 1, −B + C − D = 0.

I So A = −1/2, B = −1/2, C = −1/2, D = 0







dx






I −A + C = 0, A− B + D = 0, −A− C = 1, −B + C − D = 0.

I So A = −1/2, B = −1/2, C = −1/2, D = 0







dx






I −A + C = 0, A− B + D = 0, −A− C = 1, −B + C − D = 0.

I So A = −1/2, B = −1/2, C = −1/2, D = 0







dx






I −A + C = 0, A− B + D = 0, −A− C = 1, −B + C − D = 0.

I So A = −1/2, B = −1/2, C = −1/2, D = 0







dx






I −A + C = 0, A− B + D = 0, −A− C = 1, −B + C − D = 0.

I So A = −1/2, B = −1/2, C = −1/2, D = 0







dx






I −A + C = 0, A− B + D = 0, −A− C = 1, −B + C − D = 0.

I So A = −1/2, B = −1/2, C = −1/2, D = 0


Integrating PEO’s — II

I∫x3ex dx is a PEO of degree 3, growth 1 and frequency 0.

I∫x3ex dx = (Ax3 + Bx2 + Cx + D)ex for some A, B, C , D.

I . x3ex = ddx

((Ax3 + Bx2 + Cx + D)ex

)

= (3Ax2 + 2Bx + C)ex + (Ax3 + Bx2 + Cx + D)ex

= (Ax3 + (3A + B)x2 + (2B + C)x + (C + D))ex .

I A = 1, 3A + B = 0, 2B + C = 0, C + D = 0.

I so A = 1, B = −3, C = 6, D = −6

I so∫x3ex dx = (x3 − 3x2 + 6x − 6)ex .




I . x3ex = ddx

((Ax3 + Bx2 + Cx + D)ex

)


= (Ax3 + (3A + B)x2 + (2B + C)x + (C + D))ex .

I A = 1, 3A + B = 0, 2B + C = 0, C + D = 0.

I so A = 1, B = −3, C = 6, D = −6

I so∫x3ex dx = (x3 − 3x2 + 6x − 6)ex .




I . x3ex = ddx

((Ax3 + Bx2 + Cx + D)ex

)


= (Ax3 + (3A + B)x2 + (2B + C)x + (C + D))ex .

I A = 1, 3A + B = 0, 2B + C = 0, C + D = 0.

I so A = 1, B = −3, C = 6, D = −6

I so∫x3ex dx = (x3 − 3x2 + 6x − 6)ex .




I . x3ex = ddx

((Ax3 + Bx2 + Cx + D)ex

)


= (Ax3 + (3A + B)x2 + (2B + C)x + (C + D))ex .

I A = 1, 3A + B = 0, 2B + C = 0, C + D = 0.

I so A = 1, B = −3, C = 6, D = −6

I so∫x3ex dx = (x3 − 3x2 + 6x − 6)ex .




I . x3ex = ddx

((Ax3 + Bx2 + Cx + D)ex

)= (3Ax2 + 2Bx + C)ex + (Ax3 + Bx2 + Cx + D)ex

= (Ax3 + (3A + B)x2 + (2B + C)x + (C + D))ex .

I A = 1, 3A + B = 0, 2B + C = 0, C + D = 0.

I so A = 1, B = −3, C = 6, D = −6

I so∫x3ex dx = (x3 − 3x2 + 6x − 6)ex .




I . x3ex = ddx

((Ax3 + Bx2 + Cx + D)ex


= (Ax3 + (3A + B)x2 + (2B + C)x + (C + D))ex .

I A = 1, 3A + B = 0, 2B + C = 0, C + D = 0.

I so A = 1, B = −3, C = 6, D = −6

I so∫x3ex dx = (x3 − 3x2 + 6x − 6)ex .




I . x3ex = ddx

((Ax3 + Bx2 + Cx + D)ex


= (Ax3 + (3A + B)x2 + (2B + C)x + (C + D))ex .

I A = 1, 3A + B = 0, 2B + C = 0, C + D = 0.

I so A = 1, B = −3, C = 6, D = −6

I so∫x3ex dx = (x3 − 3x2 + 6x − 6)ex .




I . x3ex = ddx

((Ax3 + Bx2 + Cx + D)ex


= (Ax3 + (3A + B)x2 + (2B + C)x + (C + D))ex .

I A = 1, 3A + B = 0, 2B + C = 0, C + D = 0.

I so A = 1, B = −3, C = 6, D = −6

I so∫x3ex dx = (x3 − 3x2 + 6x − 6)ex .




I . x3ex = ddx

((Ax3 + Bx2 + Cx + D)ex


= (Ax3 + (3A + B)x2 + (2B + C)x + (C + D))ex .

I A = 1, 3A + B = 0, 2B + C = 0, C + D = 0.

I so A = 1, B = −3, C = 6, D = −6

I so∫x3ex dx = (x3 − 3x2 + 6x − 6)ex .

Integration by parts — I

I u = x dv/dx = ex/a

I du/dx = 1

v = a ex/a

I∫

xex/a dx = uv −∫

du

dxv dx

= axex/a −∫

a ex/a dx = axex/a − a2ex/a

I To integrate a product, call the factors u and dvdx

.

I Differentiate u to find du/dx .

I Integrate dvdx

to find v .

I Use the formula: ∫udv

dxdx = uv −

∫du

dxv dx

I This is most useful when (a) du/dx is simpler than u (eg u polynomial)and (b) v is no more complicated than dv/dx (eg dv/dx = cos(x)).


I Consider

∫xex/a dx .


I du/dx = 1

v = a ex/a

I∫


du

dxv dx

= axex/a −∫



.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx



I Consider

∫xex/a dx .


I du/dx = 1

v = a ex/a

I∫


du

dxv dx

= axex/a −∫



.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx



I Consider

∫xex/a dx .


I du/dx = 1

v = a ex/a

I∫


du

dxv dx

= axex/a −∫



.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx



I Consider

∫xex/a dx .


I du/dx = 1 v = a ex/a

I∫


du

dxv dx

= axex/a −∫



.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx



I Consider

∫xex/a dx .



I∫


du

dxv dx

= axex/a −∫



.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx



I Consider

∫xex/a dx .



I∫


du

dxv dx = axex/a −

∫a ex/a dx

= axex/a − a2ex/a


.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx



I Consider

∫xex/a dx .



I∫


du

dxv dx = axex/a −

∫a ex/a dx = axex/a − a2ex/a


.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx



I Consider

∫xex/a dx .



I∫


du

dxv dx = axex/a −

∫a ex/a dx = axex/a − a2ex/a


.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx


Integration by parts — II

I u = 1− ln(x) dv/dx = x−2

I du/dx = −x−1

v = −x−1

I∫

(1− ln(x))x−2 dx = uv −∫

du

dxv dx

= −(1− ln(x))x−1 −∫

x−2 dx

= (ln(x)− 1)x−1 + x−1 = ln(x)/x


.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx


I Consider

∫(1− ln(x))x−2 dx .

I u = 1− ln(x) dv/dx = x−2

I du/dx = −x−1

v = −x−1

I∫

(1− ln(x))x−2 dx = uv −∫

du

dxv dx

= −(1− ln(x))x−1 −∫

x−2 dx

= (ln(x)− 1)x−1 + x−1 = ln(x)/x


.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx


I Consider

∫(1− ln(x))x−2 dx .

I u = 1− ln(x) dv/dx = x−2

I du/dx = −x−1

v = −x−1

I∫

(1− ln(x))x−2 dx = uv −∫

du

dxv dx

= −(1− ln(x))x−1 −∫

x−2 dx

= (ln(x)− 1)x−1 + x−1 = ln(x)/x


.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx


I Consider

∫(1− ln(x))x−2 dx .

I u = 1− ln(x) dv/dx = x−2

I du/dx = −x−1

v = −x−1

I∫

(1− ln(x))x−2 dx = uv −∫

du

dxv dx

= −(1− ln(x))x−1 −∫

x−2 dx

= (ln(x)− 1)x−1 + x−1 = ln(x)/x


.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx


I Consider

∫(1− ln(x))x−2 dx .

I u = 1− ln(x) dv/dx = x−2

I du/dx = −x−1 v = −x−1

I∫

(1− ln(x))x−2 dx = uv −∫

du

dxv dx

= −(1− ln(x))x−1 −∫

x−2 dx

= (ln(x)− 1)x−1 + x−1 = ln(x)/x


.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx


I Consider

∫(1− ln(x))x−2 dx .

I u = 1− ln(x) dv/dx = x−2

I du/dx = −x−1 v = −x−1

I∫

(1− ln(x))x−2 dx = uv −∫

du

dxv dx

= −(1− ln(x))x−1 −∫

x−2 dx

= (ln(x)− 1)x−1 + x−1 = ln(x)/x


.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx


I Consider

∫(1− ln(x))x−2 dx .

I u = 1− ln(x) dv/dx = x−2

I du/dx = −x−1 v = −x−1

I∫

(1− ln(x))x−2 dx = uv −∫

du

dxv dx = −(1− ln(x))x−1 −

∫x−2 dx

= (ln(x)− 1)x−1 + x−1 = ln(x)/x


.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx


I Consider

∫(1− ln(x))x−2 dx .

I u = 1− ln(x) dv/dx = x−2

I du/dx = −x−1 v = −x−1

I∫

(1− ln(x))x−2 dx = uv −∫

du

dxv dx = −(1− ln(x))x−1 −

∫x−2 dx

= (ln(x)− 1)x−1 + x−1

= ln(x)/x


.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx


I Consider

∫(1− ln(x))x−2 dx .

I u = 1− ln(x) dv/dx = x−2

I du/dx = −x−1 v = −x−1

I∫

(1− ln(x))x−2 dx = uv −∫

du

dxv dx = −(1− ln(x))x−1 −

∫x−2 dx

= (ln(x)− 1)x−1 + x−1 = ln(x)/x


.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx

Integration by parts — III

I u = x dv/dx = sin(ωx)

I du/dx = 1

v = −ω−1 cos(ωx)

I∫

x sin(ωx) dx = uv −∫

du

dxv dx

= −ω−1x cos(ωx) +

∫ω−1 cos(ωx) dx

= −ω−1x cos(ωx) + ω−2 sin(ωx)


.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx


I Consider

∫x sin(ωx) dx .


I du/dx = 1


I∫


du

dxv dx





.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx


I Consider

∫x sin(ωx) dx .


I du/dx = 1


I∫


du

dxv dx





.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx


I Consider

∫x sin(ωx) dx .


I du/dx = 1


I∫


du

dxv dx





.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx


I Consider

∫x sin(ωx) dx .


I du/dx = 1 v = −ω−1 cos(ωx)

I∫


du

dxv dx





.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx


I Consider

∫x sin(ωx) dx .



I∫


du

dxv dx





.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx


I Consider

∫x sin(ωx) dx .



I∫


du

dxv dx = −ω−1x cos(ωx) +




.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx


I Consider

∫x sin(ωx) dx .



I∫


du

dxv dx = −ω−1x cos(ωx) +




.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx

Integration by parts — IV

I u = arcsin(x) dv/dx = 1

I du/dx = (1− x2)−1/2

v = x

I∫

arcsin(x).1 dx = uv −∫

du

dxv dx

= arcsin(x).x −∫

x(1− x2)−1/2 dx

= x arcsin(x) + (1− x2)1/2


.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx


I Consider

∫arcsin(x) dx .


I du/dx = (1− x2)−1/2

v = x

I∫


du

dxv dx


x(1− x2)−1/2 dx

= x arcsin(x) + (1− x2)1/2


.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx


I Consider

∫arcsin(x).1 dx .


I du/dx = (1− x2)−1/2

v = x

I∫


du

dxv dx


x(1− x2)−1/2 dx

= x arcsin(x) + (1− x2)1/2


.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx


I Consider

∫arcsin(x).1 dx .


I du/dx = (1− x2)−1/2

v = x

I∫


du

dxv dx


x(1− x2)−1/2 dx

= x arcsin(x) + (1− x2)1/2


.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx


I Consider

∫arcsin(x).1 dx .


I du/dx = (1− x2)−1/2 v = x

I∫


du

dxv dx


x(1− x2)−1/2 dx

= x arcsin(x) + (1− x2)1/2


.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx


I Consider

∫arcsin(x).1 dx .


I du/dx = (1− x2)−1/2 v = x

I∫


du

dxv dx


x(1− x2)−1/2 dx

= x arcsin(x) + (1− x2)1/2


.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx


I Consider

∫arcsin(x).1 dx .


I du/dx = (1− x2)−1/2 v = x

I∫


du

dxv dx = arcsin(x).x −

∫x(1− x2)−1/2 dx

= x arcsin(x) + (1− x2)1/2


.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx


I Consider

∫arcsin(x).1 dx .


I du/dx = (1− x2)−1/2 v = x

I∫


du

dxv dx = arcsin(x).x −

∫x(1− x2)−1/2 dx

= x arcsin(x) + (1− x2)1/2


.


I Integrate dvdx

to find v .


dxdx = uv −

∫du

dxv dx

Integration by substitution — I

I Put u = cos(x)

, so du/dx = − sin(x), so dx = −du/ sin(x)

I .

∫sin(x)

cos(x)ndx =

∫sin(x)

un

−dusin(x)

= −∫

u−n du

= u1−n/(n − 1) =cos(x)1−n

n − 1

I To find∫f (x) dx , pick out some part of f (x) and call it u.

I Find du/dx

, and rearrange to express dx in terms of x and du.

I Rewrite the integral in terms of u and du.

I Evaluate the integral

, then rewrite the result in terms of x .


I Consider

∫sin(x)

cos(x)ndx .

I Put u = cos(x)


I .

∫sin(x)

cos(x)ndx =

∫sin(x)

un

−dusin(x)

= −∫

u−n du

= u1−n/(n − 1) =cos(x)1−n

n − 1


I Find du/dx






I Consider

∫sin(x)

cos(x)ndx .

I Put u = cos(x)


I .

∫sin(x)

cos(x)ndx =

∫sin(x)

un

−dusin(x)

= −∫

u−n du

= u1−n/(n − 1) =cos(x)1−n

n − 1


I Find du/dx






I Consider

∫sin(x)

cos(x)ndx .

I Put u = cos(x), so du/dx = − sin(x)

, so dx = −du/ sin(x)

I .

∫sin(x)

cos(x)ndx =

∫sin(x)

un

−dusin(x)

= −∫

u−n du

= u1−n/(n − 1) =cos(x)1−n

n − 1


I Find du/dx






I Consider

∫sin(x)

cos(x)ndx .

I Put u = cos(x), so du/dx = − sin(x), so dx = −du/ sin(x)

I .

∫sin(x)

cos(x)ndx =

∫sin(x)

un

−dusin(x)

= −∫

u−n du

= u1−n/(n − 1) =cos(x)1−n

n − 1


I Find du/dx , and rearrange to express dx in terms of x and du.





I Consider

∫sin(x)

cos(x)ndx .


I .

∫sin(x)

cos(x)ndx =

∫sin(x)

un

−dusin(x)

= −∫

u−n du

= u1−n/(n − 1) =cos(x)1−n

n − 1







I Consider

∫sin(x)

cos(x)ndx .


I .

∫sin(x)

cos(x)ndx =

∫sin(x)

un

−dusin(x)

= −∫

u−n du

= u1−n/(n − 1) =cos(x)1−n

n − 1







I Consider

∫sin(x)

cos(x)ndx .


I .

∫sin(x)

cos(x)ndx =

∫sin(x)

un

−dusin(x)

= −∫

u−n du

= u1−n/(n − 1)

=cos(x)1−n

n − 1







I Consider

∫sin(x)

cos(x)ndx .


I .

∫sin(x)

cos(x)ndx =

∫sin(x)

un

−dusin(x)

= −∫

u−n du

= u1−n/(n − 1) =cos(x)1−n

n − 1




I Evaluate the integral, then rewrite the result in terms of x .

Integration by substitution — II

I Put u = −4x2

, so du/dx = −8x , so dx = −du/(8x)

I .

∫xe−4x2

dx =

∫−xeu du

8x

= −1

8

∫eu du

= −eu/8 = −e−4x2

/8


I Find du/dx






I Consider

∫xe−4x2

dx .

I Put u = −4x2


I .

∫xe−4x2

dx =

∫−xeu du

8x

= −1

8

∫eu du

= −eu/8 = −e−4x2

/8


I Find du/dx






I Consider

∫xe−4x2

dx .

I Put u = −4x2


I .

∫xe−4x2

dx =

∫−xeu du

8x

= −1

8

∫eu du

= −eu/8 = −e−4x2

/8


I Find du/dx






I Consider

∫xe−4x2

dx .

I Put u = −4x2, so du/dx = −8x

, so dx = −du/(8x)

I .

∫xe−4x2

dx =

∫−xeu du

8x

= −1

8

∫eu du

= −eu/8 = −e−4x2

/8


I Find du/dx






I Consider

∫xe−4x2

dx .

I Put u = −4x2, so du/dx = −8x , so dx = −du/(8x)

I .

∫xe−4x2

dx =

∫−xeu du

8x

= −1

8

∫eu du

= −eu/8 = −e−4x2

/8







I Consider

∫xe−4x2

dx .


I .

∫xe−4x2

dx =

∫−xeu du

8x

= −1

8

∫eu du

= −eu/8 = −e−4x2

/8







I Consider

∫xe−4x2

dx .


I .

∫xe−4x2

dx =

∫−xeu du

8x= −1

8

∫eu du

= −eu/8 = −e−4x2

/8







I Consider

∫xe−4x2

dx .


I .

∫xe−4x2

dx =

∫−xeu du

8x= −1

8

∫eu du

= −eu/8

= −e−4x2

/8







I Consider

∫xe−4x2

dx .


I .

∫xe−4x2

dx =

∫−xeu du

8x= −1

8

∫eu du

= −eu/8 = −e−4x2

/8





Integration by substitution — III

I Put u = 2x + 1

, so du/dx = 2, so dx = du/2

I .

∫dx

4x2 + 4x + 2=

∫du/2

u2 + 1

= arctan(u)/2 = arctan(2x + 1)/2


I Find du/dx






I Consider

∫dx

4x2 + 4x + 2.

I Put u = 2x + 1


I .

∫dx

4x2 + 4x + 2=

∫du/2

u2 + 1



I Find du/dx






I Consider

∫dx

4x2 + 4x + 2=

∫dx

(2x + 1)2 + 1.

I Put u = 2x + 1


I .

∫dx

4x2 + 4x + 2=

∫du/2

u2 + 1



I Find du/dx






I Consider

∫dx

4x2 + 4x + 2=

∫dx

(2x + 1)2 + 1.

I Put u = 2x + 1, so du/dx = 2

, so dx = du/2

I .

∫dx

4x2 + 4x + 2=

∫du/2

u2 + 1



I Find du/dx






I Consider

∫dx

4x2 + 4x + 2=

∫dx

(2x + 1)2 + 1.

I Put u = 2x + 1, so du/dx = 2, so dx = du/2

I .

∫dx

4x2 + 4x + 2=

∫du/2

u2 + 1








I Consider

∫dx

4x2 + 4x + 2=

∫dx

(2x + 1)2 + 1.


I .

∫dx

4x2 + 4x + 2=

∫du/2

u2 + 1








I Consider

∫dx

4x2 + 4x + 2=

∫dx

(2x + 1)2 + 1.


I .

∫dx

4x2 + 4x + 2=

∫du/2

u2 + 1

= arctan(u)/2

= arctan(2x + 1)/2







I Consider

∫dx

4x2 + 4x + 2=

∫dx

(2x + 1)2 + 1.


I .

∫dx

4x2 + 4x + 2=

∫du/2

u2 + 1






Integration by substitution — IV

I Consider

∫dx√x − x2

.

I Put x = t2

, so dx/dt = 2t, so dx = 2t dt√x − x2 =

√t2 − t4 = t

√1− t2∫

dx√x − x2

=

∫2t dt

t√

1− t2= 2

∫dt√

1− t2

= 2 arcsin(t) = 2 arcsin(√x)

I To find∫f (x) dx , put x equal to some function of t.

I Find dx/dt

, and rearrange to express dx in terms of t and dt.

I Rewrite the integral in terms of t and dt.




I Consider

∫dx√x − x2

.

I Put x = t2


√t2 − t4 = t

√1− t2∫

dx√x − x2

=

∫2t dt

t√

1− t2= 2

∫dt√

1− t2



I Find dx/dt






I Consider

∫dx√x − x2

.

I Put x = t2


√t2 − t4 = t

√1− t2∫

dx√x − x2

=

∫2t dt

t√

1− t2= 2

∫dt√

1− t2



I Find dx/dt






I Consider

∫dx√x − x2

.

I Put x = t2, so dx/dt = 2t

, so dx = 2t dt√x − x2 =

√t2 − t4 = t

√1− t2∫

dx√x − x2

=

∫2t dt

t√

1− t2= 2

∫dt√

1− t2



I Find dx/dt






I Consider

∫dx√x − x2

.

I Put x = t2, so dx/dt = 2t, so dx = 2t dt

√x − x2 =

√t2 − t4 = t

√1− t2∫

dx√x − x2

=

∫2t dt

t√

1− t2= 2

∫dt√

1− t2



I Find dx/dt, and rearrange to express dx in terms of t and dt.





I Consider

∫dx√x − x2

.

I Put x = t2, so dx/dt = 2t, so dx = 2t dt√x − x2 =

√t2 − t4 = t

√1− t2

∫dx√x − x2

=

∫2t dt

t√

1− t2= 2

∫dt√

1− t2








I Consider

∫dx√x − x2

.


√t2 − t4 = t

√1− t2∫

dx√x − x2

=

∫2t dt

t√

1− t2

= 2

∫dt√

1− t2








I Consider

∫dx√x − x2

.


√t2 − t4 = t

√1− t2∫

dx√x − x2

=

∫2t dt

t√

1− t2= 2

∫dt√

1− t2








I Consider

∫dx√x − x2

.


√t2 − t4 = t

√1− t2∫

dx√x − x2

=

∫2t dt

t√

1− t2= 2

∫dt√

1− t2

= 2 arcsin(t)

= 2 arcsin(√x)







I Consider

∫dx√x − x2

.


√t2 − t4 = t

√1− t2∫

dx√x − x2

=

∫2t dt

t√

1− t2= 2

∫dt√

1− t2






Integration by substitution — V

I Consider

∫log(x)2 dx .

I Put x = et

, so dx/dt = et , so dx = et dt∫log(x)2 dx =

∫log(et)2et dt =

∫t2et dt

= (t2 − 2t + 2)et = (log(x)2 − 2 log(x) + 2)x


I Find dx/dt






I Consider

∫log(x)2 dx .

I Put x = et


∫log(et)2et dt =

∫t2et dt

= (t2 − 2t + 2)et = (log(x)2 − 2 log(x) + 2)x


I Find dx/dt






I Consider

∫log(x)2 dx .

I Put x = et


∫log(et)2et dt =

∫t2et dt

= (t2 − 2t + 2)et = (log(x)2 − 2 log(x) + 2)x


I Find dx/dt






I Consider

∫log(x)2 dx .

I Put x = et , so dx/dt = et

, so dx = et dt∫log(x)2 dx =

∫log(et)2et dt =

∫t2et dt

= (t2 − 2t + 2)et = (log(x)2 − 2 log(x) + 2)x


I Find dx/dt






I Consider

∫log(x)2 dx .

I Put x = et , so dx/dt = et , so dx = et dt

∫log(x)2 dx =

∫log(et)2et dt =

∫t2et dt

= (t2 − 2t + 2)et = (log(x)2 − 2 log(x) + 2)x







I Consider

∫log(x)2 dx .

I Put x = et , so dx/dt = et , so dx = et dt∫log(x)2 dx =

∫log(et)2et dt

=

∫t2et dt

= (t2 − 2t + 2)et = (log(x)2 − 2 log(x) + 2)x







I Consider

∫log(x)2 dx .


∫log(et)2et dt =

∫t2et dt

= (t2 − 2t + 2)et = (log(x)2 − 2 log(x) + 2)x







I Consider

∫log(x)2 dx .


∫log(et)2et dt =

∫t2et dt

= (t2 − 2t + 2)et

= (log(x)2 − 2 log(x) + 2)x







I Consider

∫log(x)2 dx .


∫log(et)2et dt =

∫t2et dt

= (t2 − 2t + 2)et = (log(x)2 − 2 log(x) + 2)x





Examples I

I∫

tan(x) dx

=

∫sin(x)

cos(x)dx = −

∫cos′(x)

cos(x)dx = − log(cos(x)).

I Consider∫x2 tan(x3) dx .

Put u = x3, so du = 3x2 dx , so dx = du/(3x2).∫x2 tan(x3) dx =

∫x2 tan(u)

du

3x2=

1

3

∫tan(u) du = − log(cos(u))/3

= − log(cos(x3))/3

I Consider∫xe√

x dx .

Put t =√x , so x = t2, so dx = 2t dt.∫

xe√x dx =

∫t2et .2t dt = 2

∫t3et dt = 2(t3 − 3t2 + 6t − 6)et

= (2x3/2 − 6x + 12x1/2 − 12)e√

x

Examples I

I∫

tan(x) dx

=

∫sin(x)

cos(x)dx = −

∫cos′(x)




∫x2 tan(u)

du

3x2=

1

3


= − log(cos(x3))/3

I Consider∫xe√

x dx .


xe√x dx =

∫t2et .2t dt = 2

∫t3et dt = 2(t3 − 3t2 + 6t − 6)et

= (2x3/2 − 6x + 12x1/2 − 12)e√x

Examples I

I∫

tan(x) dx =

∫sin(x)

cos(x)dx

= −∫

cos′(x)




∫x2 tan(u)

du

3x2=

1

3


= − log(cos(x3))/3

I Consider∫xe√

x dx .


xe√x dx =

∫t2et .2t dt = 2

∫t3et dt = 2(t3 − 3t2 + 6t − 6)et

= (2x3/2 − 6x + 12x1/2 − 12)e√x

Examples I

I∫

tan(x) dx =

∫sin(x)

cos(x)dx = −

∫cos′(x)

cos(x)dx

= − log(cos(x)).



∫x2 tan(u)

du

3x2=

1

3


= − log(cos(x3))/3

I Consider∫xe√

x dx .


xe√x dx =

∫t2et .2t dt = 2

∫t3et dt = 2(t3 − 3t2 + 6t − 6)et

= (2x3/2 − 6x + 12x1/2 − 12)e√x

Examples I

I∫

tan(x) dx =

∫sin(x)

cos(x)dx = −

∫cos′(x)




∫x2 tan(u)

du

3x2=

1

3


= − log(cos(x3))/3

I Consider∫xe√

x dx .


xe√x dx =

∫t2et .2t dt = 2

∫t3et dt = 2(t3 − 3t2 + 6t − 6)et

= (2x3/2 − 6x + 12x1/2 − 12)e√

x

Examples I

I∫

tan(x) dx =

∫sin(x)

cos(x)dx = −

∫cos′(x)




∫x2 tan(u)

du

3x2=

1

3


= − log(cos(x3))/3

I Consider∫xe√

x dx .


xe√x dx =

∫t2et .2t dt = 2

∫t3et dt = 2(t3 − 3t2 + 6t − 6)et

= (2x3/2 − 6x + 12x1/2 − 12)e√

x

Examples I

I∫

tan(x) dx =

∫sin(x)

cos(x)dx = −

∫cos′(x)


I Consider∫x2 tan(x3) dx . Put u = x3, so du = 3x2 dx , so dx = du/(3x2).

∫x2 tan(x3) dx =

∫x2 tan(u)

du

3x2=

1

3


= − log(cos(x3))/3

I Consider∫xe√

x dx .


xe√x dx =

∫t2et .2t dt = 2

∫t3et dt = 2(t3 − 3t2 + 6t − 6)et

= (2x3/2 − 6x + 12x1/2 − 12)e√

x

Examples I

I∫

tan(x) dx =

∫sin(x)

cos(x)dx = −

∫cos′(x)


I Consider∫x2 tan(x3) dx . Put u = x3, so du = 3x2 dx , so dx = du/(3x2).∫

x2 tan(x3) dx =

∫x2 tan(u)

du

3x2

=1

3


= − log(cos(x3))/3

I Consider∫xe√

x dx .


xe√x dx =

∫t2et .2t dt = 2

∫t3et dt = 2(t3 − 3t2 + 6t − 6)et

= (2x3/2 − 6x + 12x1/2 − 12)e√x

Examples I

I∫

tan(x) dx =

∫sin(x)

cos(x)dx = −

∫cos′(x)



x2 tan(x3) dx =

∫x2 tan(u)

du

3x2=

1

3

∫tan(u) du

= − log(cos(u))/3

= − log(cos(x3))/3

I Consider∫xe√

x dx .


xe√x dx =

∫t2et .2t dt = 2

∫t3et dt = 2(t3 − 3t2 + 6t − 6)et

= (2x3/2 − 6x + 12x1/2 − 12)e√x

Examples I

I∫

tan(x) dx =

∫sin(x)

cos(x)dx = −

∫cos′(x)



x2 tan(x3) dx =

∫x2 tan(u)

du

3x2=

1

3


= − log(cos(x3))/3

I Consider∫xe√

x dx .


xe√x dx =

∫t2et .2t dt = 2

∫t3et dt = 2(t3 − 3t2 + 6t − 6)et

= (2x3/2 − 6x + 12x1/2 − 12)e√x

Examples I

I∫

tan(x) dx =

∫sin(x)

cos(x)dx = −

∫cos′(x)



x2 tan(x3) dx =

∫x2 tan(u)

du

3x2=

1

3


= − log(cos(x3))/3

I Consider∫xe√

x dx .


xe√x dx =

∫t2et .2t dt = 2

∫t3et dt = 2(t3 − 3t2 + 6t − 6)et

= (2x3/2 − 6x + 12x1/2 − 12)e√x

Examples I

I∫

tan(x) dx =

∫sin(x)

cos(x)dx = −

∫cos′(x)



x2 tan(x3) dx =

∫x2 tan(u)

du

3x2=

1

3


= − log(cos(x3))/3

I Consider∫xe√x dx .


xe√x dx =

∫t2et .2t dt = 2

∫t3et dt = 2(t3 − 3t2 + 6t − 6)et

= (2x3/2 − 6x + 12x1/2 − 12)e√

x

Examples I

I∫

tan(x) dx =

∫sin(x)

cos(x)dx = −

∫cos′(x)



x2 tan(x3) dx =

∫x2 tan(u)

du

3x2=

1

3


= − log(cos(x3))/3

I Consider∫xe√x dx . Put t =

√x , so x = t2, so dx = 2t dt.

∫xe√x dx =

∫t2et .2t dt = 2

∫t3et dt = 2(t3 − 3t2 + 6t − 6)et

= (2x3/2 − 6x + 12x1/2 − 12)e√

x

Examples I

I∫

tan(x) dx =

∫sin(x)

cos(x)dx = −

∫cos′(x)



x2 tan(x3) dx =

∫x2 tan(u)

du

3x2=

1

3


= − log(cos(x3))/3


√x , so x = t2, so dx = 2t dt.∫

xe√x dx =

∫t2et .2t dt

= 2

∫t3et dt = 2(t3 − 3t2 + 6t − 6)et

= (2x3/2 − 6x + 12x1/2 − 12)e√

x

Examples I

I∫

tan(x) dx =

∫sin(x)

cos(x)dx = −

∫cos′(x)



x2 tan(x3) dx =

∫x2 tan(u)

du

3x2=

1

3


= − log(cos(x3))/3


√x , so x = t2, so dx = 2t dt.∫

xe√x dx =

∫t2et .2t dt = 2

∫t3et dt

= 2(t3 − 3t2 + 6t − 6)et

= (2x3/2 − 6x + 12x1/2 − 12)e√

x

Examples I

I∫

tan(x) dx =

∫sin(x)

cos(x)dx = −

∫cos′(x)



x2 tan(x3) dx =

∫x2 tan(u)

du

3x2=

1

3


= − log(cos(x3))/3


√x , so x = t2, so dx = 2t dt.∫

xe√x dx =

∫t2et .2t dt = 2

∫t3et dt = 2(t3 − 3t2 + 6t − 6)et

= (2x3/2 − 6x + 12x1/2 − 12)e√

x

Examples I

I∫

tan(x) dx =

∫sin(x)

cos(x)dx = −

∫cos′(x)



x2 tan(x3) dx =

∫x2 tan(u)

du

3x2=

1

3


= − log(cos(x3))/3


√x , so x = t2, so dx = 2t dt.∫

xe√x dx =

∫t2et .2t dt = 2

∫t3et dt = 2(t3 − 3t2 + 6t − 6)et

= (2x3/2 − 6x + 12x1/2 − 12)e√

x

Examples II

I .

∫(2(x2 + 1)ex)2 dx

=

∫(4x4 + 8x2 + 4)e2x dx

= (Ax4 + Bx3 + Cx2 + Dx + E)e2x

(4x4 + 8x2 + 4)e2x =d

dx((Ax4 + Bx3 + Cx2 + Dx + E)e2x)

= (4Ax3 + 3Bx2 + 2Cx + D)e2x +

(Ax4 + Bx3 + Cx2 + Dx + E).2e2x

= e2x(2Ax4 + (4A + 2B)x3 + (3B + 2C)x2 +

(2C + 2D)x + (D + 2E))

So 4 = 2A, 0 = 4A + 2B, 8 = 3B + 2C , 0 = 2C + 2D, 4 = D + 2ESo A = 2, B = −4, C = 10, D = −10, E = 7∫

(2(x2 + 1)ex)2 dx = (2x4 − 4x3 + 10x2 − 10x + 7)e2x .

Examples II

I .

∫(2(x2 + 1)ex)2 dx

=

∫(4x4 + 8x2 + 4)e2x dx

= (Ax4 + Bx3 + Cx2 + Dx + E)e2x

(4x4 + 8x2 + 4)e2x =d

dx((Ax4 + Bx3 + Cx2 + Dx + E)e2x)

= (4Ax3 + 3Bx2 + 2Cx + D)e2x +

(Ax4 + Bx3 + Cx2 + Dx + E).2e2x

= e2x(2Ax4 + (4A + 2B)x3 + (3B + 2C)x2 +

(2C + 2D)x + (D + 2E))


(2(x2 + 1)ex)2 dx = (2x4 − 4x3 + 10x2 − 10x + 7)e2x .

Examples II

I .

∫(2(x2 + 1)ex)2 dx =

∫(4x4 + 8x2 + 4)e2x dx

= (Ax4 + Bx3 + Cx2 + Dx + E)e2x

(4x4 + 8x2 + 4)e2x =d

dx((Ax4 + Bx3 + Cx2 + Dx + E)e2x)

= (4Ax3 + 3Bx2 + 2Cx + D)e2x +

(Ax4 + Bx3 + Cx2 + Dx + E).2e2x

= e2x(2Ax4 + (4A + 2B)x3 + (3B + 2C)x2 +

(2C + 2D)x + (D + 2E))


(2(x2 + 1)ex)2 dx = (2x4 − 4x3 + 10x2 − 10x + 7)e2x .

Examples II

I .

∫(2(x2 + 1)ex)2 dx =

∫(4x4 + 8x2 + 4)e2x dx

= (Ax4 + Bx3 + Cx2 + Dx + E)e2x

(4x4 + 8x2 + 4)e2x =d

dx((Ax4 + Bx3 + Cx2 + Dx + E)e2x)

= (4Ax3 + 3Bx2 + 2Cx + D)e2x +

(Ax4 + Bx3 + Cx2 + Dx + E).2e2x

= e2x(2Ax4 + (4A + 2B)x3 + (3B + 2C)x2 +

(2C + 2D)x + (D + 2E))


(2(x2 + 1)ex)2 dx = (2x4 − 4x3 + 10x2 − 10x + 7)e2x .

Examples II

I .

∫(2(x2 + 1)ex)2 dx =

∫(4x4 + 8x2 + 4)e2x dx

= (Ax4 + Bx3 + Cx2 + Dx + E)e2x

(4x4 + 8x2 + 4)e2x =d

dx((Ax4 + Bx3 + Cx2 + Dx + E)e2x)

= (4Ax3 + 3Bx2 + 2Cx + D)e2x +

(Ax4 + Bx3 + Cx2 + Dx + E).2e2x

= e2x(2Ax4 + (4A + 2B)x3 + (3B + 2C)x2 +

(2C + 2D)x + (D + 2E))


(2(x2 + 1)ex)2 dx = (2x4 − 4x3 + 10x2 − 10x + 7)e2x .

Examples II

I .

∫(2(x2 + 1)ex)2 dx =

∫(4x4 + 8x2 + 4)e2x dx

= (Ax4 + Bx3 + Cx2 + Dx + E)e2x

(4x4 + 8x2 + 4)e2x =d

dx((Ax4 + Bx3 + Cx2 + Dx + E)e2x)

= (4Ax3 + 3Bx2 + 2Cx + D)e2x +

(Ax4 + Bx3 + Cx2 + Dx + E).2e2x

= e2x(2Ax4 + (4A + 2B)x3 + (3B + 2C)x2 +

(2C + 2D)x + (D + 2E))


(2(x2 + 1)ex)2 dx = (2x4 − 4x3 + 10x2 − 10x + 7)e2x .

Examples II

I .

∫(2(x2 + 1)ex)2 dx =

∫(4x4 + 8x2 + 4)e2x dx

= (Ax4 + Bx3 + Cx2 + Dx + E)e2x

(4x4 + 8x2 + 4)e2x =d

dx((Ax4 + Bx3 + Cx2 + Dx + E)e2x)

= (4Ax3 + 3Bx2 + 2Cx + D)e2x +

(Ax4 + Bx3 + Cx2 + Dx + E).2e2x

= e2x(2Ax4 + (4A + 2B)x3 + (3B + 2C)x2 +

(2C + 2D)x + (D + 2E))


(2(x2 + 1)ex)2 dx = (2x4 − 4x3 + 10x2 − 10x + 7)e2x .

Examples II

I .

∫(2(x2 + 1)ex)2 dx =

∫(4x4 + 8x2 + 4)e2x dx

= (Ax4 + Bx3 + Cx2 + Dx + E)e2x

(4x4 + 8x2 + 4)e2x =d

dx((Ax4 + Bx3 + Cx2 + Dx + E)e2x)

= (4Ax3 + 3Bx2 + 2Cx + D)e2x +

(Ax4 + Bx3 + Cx2 + Dx + E).2e2x

= e2x(2Ax4 + (4A + 2B)x3 + (3B + 2C)x2 +

(2C + 2D)x + (D + 2E))

So 4 = 2A, 0 = 4A + 2B, 8 = 3B + 2C , 0 = 2C + 2D, 4 = D + 2E

So A = 2, B = −4, C = 10, D = −10, E = 7∫(2(x2 + 1)ex)2 dx = (2x4 − 4x3 + 10x2 − 10x + 7)e2x .

Examples II

I .

∫(2(x2 + 1)ex)2 dx =

∫(4x4 + 8x2 + 4)e2x dx

= (Ax4 + Bx3 + Cx2 + Dx + E)e2x

(4x4 + 8x2 + 4)e2x =d

dx((Ax4 + Bx3 + Cx2 + Dx + E)e2x)

= (4Ax3 + 3Bx2 + 2Cx + D)e2x +

(Ax4 + Bx3 + Cx2 + Dx + E).2e2x

= e2x(2Ax4 + (4A + 2B)x3 + (3B + 2C)x2 +

(2C + 2D)x + (D + 2E))

So 4 = 2A, 0 = 4A + 2B, 8 = 3B + 2C , 0 = 2C + 2D, 4 = D + 2ESo A = 2, B = −4, C = 10, D = −10, E = 7

∫(2(x2 + 1)ex)2 dx = (2x4 − 4x3 + 10x2 − 10x + 7)e2x .

Examples II

I .

∫(2(x2 + 1)ex)2 dx =

∫(4x4 + 8x2 + 4)e2x dx

= (Ax4 + Bx3 + Cx2 + Dx + E)e2x

(4x4 + 8x2 + 4)e2x =d

dx((Ax4 + Bx3 + Cx2 + Dx + E)e2x)

= (4Ax3 + 3Bx2 + 2Cx + D)e2x +

(Ax4 + Bx3 + Cx2 + Dx + E).2e2x

= e2x(2Ax4 + (4A + 2B)x3 + (3B + 2C)x2 +

(2C + 2D)x + (D + 2E))


(2(x2 + 1)ex)2 dx = (2x4 − 4x3 + 10x2 − 10x + 7)e2x .

Examples III

I .

∫1 + cosh(x) + cosh(x)2 dx

=

∫1 +

ex + e−x

2+

(ex + e−x

2

)2

dx

=1

4

∫4 + 2ex + 2e−x + e2x + 2 + e−2x dx

=1

4

(6x + 2ex − 2e−x + 1

2e2x − 1

2e−2x

)=

3

2x +

ex − e−x

2+

1

4

e2x − e−2x

2

=3

2x + sinh(x) +

1

4sinh(2x).

Examples III

I .

∫1 + cosh(x) + cosh(x)2 dx

=

∫1 +

ex + e−x

2+

(ex + e−x

2

)2

dx

=1

4

∫4 + 2ex + 2e−x + e2x + 2 + e−2x dx

=1

4

(6x + 2ex − 2e−x + 1

2e2x − 1

2e−2x

)=

3

2x +

ex − e−x

2+

1

4

e2x − e−2x

2

=3

2x + sinh(x) +

1

4sinh(2x).

Examples III

I .

∫1 + cosh(x) + cosh(x)2 dx =

∫1 +

ex + e−x

2+

(ex + e−x

2

)2

dx

=1

4

∫4 + 2ex + 2e−x + e2x + 2 + e−2x dx

=1

4

(6x + 2ex − 2e−x + 1

2e2x − 1

2e−2x

)=

3

2x +

ex − e−x

2+

1

4

e2x − e−2x

2

=3

2x + sinh(x) +

1

4sinh(2x).

Examples III

I .


∫1 +

ex + e−x

2+

(ex + e−x

2

)2

dx

=1

4

∫4 + 2ex + 2e−x + e2x + 2 + e−2x dx

=1

4

(6x + 2ex − 2e−x + 1

2e2x − 1

2e−2x

)=

3

2x +

ex − e−x

2+

1

4

e2x − e−2x

2

=3

2x + sinh(x) +

1

4sinh(2x).

Examples III

I .


∫1 +

ex + e−x

2+

(ex + e−x

2

)2

dx

=1

4

∫4 + 2ex + 2e−x + e2x + 2 + e−2x dx

=1

4

(6x + 2ex − 2e−x + 1

2e2x − 1

2e−2x

)

=3

2x +

ex − e−x

2+

1

4

e2x − e−2x

2

=3

2x + sinh(x) +

1

4sinh(2x).

Examples III

I .


∫1 +

ex + e−x

2+

(ex + e−x

2

)2

dx

=1

4

∫4 + 2ex + 2e−x + e2x + 2 + e−2x dx

=1

4

(6x + 2ex − 2e−x + 1

2e2x − 1

2e−2x

)=

3

2x +

ex − e−x

2+

1

4

e2x − e−2x

2

=3

2x + sinh(x) +

1

4sinh(2x).

Examples III

I .


∫1 +

ex + e−x

2+

(ex + e−x

2

)2

dx

=1

4

∫4 + 2ex + 2e−x + e2x + 2 + e−2x dx

=1

4

(6x + 2ex − 2e−x + 1

2e2x − 1

2e−2x

)=

3

2x +

ex − e−x

2+

1

4

e2x − e−2x

2

=3

2x + sinh(x) +

1

4sinh(2x).

Examples IV

I To show that

∫dx

cos(x)= log

(1 + sin(x)

cos(x)

):

d

dx

(1 + sin(x)

cos(x)

)=

cos(x). cos(x)− (1 + sin(x))(− sin(x))

cos(x)2

=cos(x)2 + sin(x)2 + sin(x)

cos(x)2=

1 + sin(x)

cos(x)2

d

dxlog

(1 + sin(x)

cos(x)

)=

(1 + sin(x)

cos(x)

)−1d

dx

(1 + sin(x)

cos(x)

)=

cos(x)

1 + sin(x)

1 + sin(x)

cos(x)2=

1

cos(x)

Examples IV

I To show that

∫dx

cos(x)= log

(1 + sin(x)

cos(x)

):

d

dx

(1 + sin(x)

cos(x)

)=


cos(x)2


cos(x)2=

1 + sin(x)

cos(x)2

d

dxlog

(1 + sin(x)

cos(x)

)=

(1 + sin(x)

cos(x)

)−1d

dx

(1 + sin(x)

cos(x)

)=

cos(x)

1 + sin(x)

1 + sin(x)

cos(x)2=

1

cos(x)

Examples IV

I To show that

∫dx

cos(x)= log

(1 + sin(x)

cos(x)

):

d

dx

(1 + sin(x)

cos(x)

)=


cos(x)2


cos(x)2=

1 + sin(x)

cos(x)2

d

dxlog

(1 + sin(x)

cos(x)

)=

(1 + sin(x)

cos(x)

)−1d

dx

(1 + sin(x)

cos(x)

)=

cos(x)

1 + sin(x)

1 + sin(x)

cos(x)2=

1

cos(x)

Examples IV

I To show that

∫dx

cos(x)= log

(1 + sin(x)

cos(x)

):

d

dx

(1 + sin(x)

cos(x)

)=


cos(x)2


cos(x)2

=1 + sin(x)

cos(x)2

d

dxlog

(1 + sin(x)

cos(x)

)=

(1 + sin(x)

cos(x)

)−1d

dx

(1 + sin(x)

cos(x)

)=

cos(x)

1 + sin(x)

1 + sin(x)

cos(x)2=

1

cos(x)

Examples IV

I To show that

∫dx

cos(x)= log

(1 + sin(x)

cos(x)

):

d

dx

(1 + sin(x)

cos(x)

)=


cos(x)2


cos(x)2=

1 + sin(x)

cos(x)2

d

dxlog

(1 + sin(x)

cos(x)

)=

(1 + sin(x)

cos(x)

)−1d

dx

(1 + sin(x)

cos(x)

)=

cos(x)

1 + sin(x)

1 + sin(x)

cos(x)2=

1

cos(x)

Examples IV

I To show that

∫dx

cos(x)= log

(1 + sin(x)

cos(x)

):

d

dx

(1 + sin(x)

cos(x)

)=


cos(x)2


cos(x)2=

1 + sin(x)

cos(x)2

d

dxlog

(1 + sin(x)

cos(x)

)=

(1 + sin(x)

cos(x)

)−1d

dx

(1 + sin(x)

cos(x)

)

=cos(x)

1 + sin(x)

1 + sin(x)

cos(x)2=

1

cos(x)

Examples IV

I To show that

∫dx

cos(x)= log

(1 + sin(x)

cos(x)

):

d

dx

(1 + sin(x)

cos(x)

)=


cos(x)2


cos(x)2=

1 + sin(x)

cos(x)2

d

dxlog

(1 + sin(x)

cos(x)

)=

(1 + sin(x)

cos(x)

)−1d

dx

(1 + sin(x)

cos(x)

)=

cos(x)

1 + sin(x)

1 + sin(x)

cos(x)2

=1

cos(x)

Examples IV

I To show that

∫dx

cos(x)= log

(1 + sin(x)

cos(x)

):

d

dx

(1 + sin(x)

cos(x)

)=


cos(x)2


cos(x)2=

1 + sin(x)

cos(x)2

d

dxlog

(1 + sin(x)

cos(x)

)=

(1 + sin(x)

cos(x)

)−1d

dx

(1 + sin(x)

cos(x)

)=

cos(x)

1 + sin(x)

1 + sin(x)

cos(x)2=

1

cos(x)

Examples V

I .

∫8x sin(x) cos(x) dx

=

∫4x sin(2x) dx

= −2x cos(2x) +

∫2 cos(2x) dx

= −2x cos(2x) + sin(2x).

I Consider

∫10e−x sin(x)2 dx

=

∫5e−x dx +

∫−5e−x cos(2x) dx .∫

−5e−x cos(2x) dx = e−x(A cos(2x) + B sin(2x))

− 5e−x cos(2x) = e−x((2B − A) cos(2x)− (2A + B) sin(2x))

A = 1, B = −2∫10e−x sin(x)2 dx = −5e−x + e−x cos(2x)− 2e−x sin(2x).

Examples V

I .

∫8x sin(x) cos(x) dx

=

∫4x sin(2x) dx

= −2x cos(2x) +

∫2 cos(2x) dx

= −2x cos(2x) + sin(2x).

I Consider


=

∫5e−x dx +





Examples V

I .

∫8x sin(x) cos(x) dx =

∫4x sin(2x) dx

= −2x cos(2x) +

∫2 cos(2x) dx

= −2x cos(2x) + sin(2x).

I Consider


=

∫5e−x dx +





Examples V

I .


∫4x sin(2x) dx

= −2x cos(2x) +

∫2 cos(2x) dx

= −2x cos(2x) + sin(2x).

I Consider


=

∫5e−x dx +





Examples V

I .


∫4x sin(2x) dx

= −2x cos(2x) +

∫2 cos(2x) dx

= −2x cos(2x) + sin(2x).

I Consider


=

∫5e−x dx +





Examples V

I .


∫4x sin(2x) dx

= −2x cos(2x) +

∫2 cos(2x) dx

= −2x cos(2x) + sin(2x).

I Consider


=

∫5e−x dx +





Examples V

I .


∫4x sin(2x) dx

= −2x cos(2x) +

∫2 cos(2x) dx

= −2x cos(2x) + sin(2x).

I Consider

∫10e−x sin(x)2 dx =

∫5e−x dx +

∫−5e−x cos(2x) dx .

∫−5e−x cos(2x) dx = e−x(A cos(2x) + B sin(2x))



Examples V

I .


∫4x sin(2x) dx

= −2x cos(2x) +

∫2 cos(2x) dx

= −2x cos(2x) + sin(2x).

I Consider


∫5e−x dx +





Examples V

I .


∫4x sin(2x) dx

= −2x cos(2x) +

∫2 cos(2x) dx

= −2x cos(2x) + sin(2x).

I Consider


∫5e−x dx +





Examples V

I .


∫4x sin(2x) dx

= −2x cos(2x) +

∫2 cos(2x) dx

= −2x cos(2x) + sin(2x).

I Consider


∫5e−x dx +




A = 1, B = −2

∫10e−x sin(x)2 dx = −5e−x + e−x cos(2x)− 2e−x sin(2x).

Examples V

I .


∫4x sin(2x) dx

= −2x cos(2x) +

∫2 cos(2x) dx

= −2x cos(2x) + sin(2x).

I Consider


∫5e−x dx +





Taylor series

ex = exp(x) = 1 + x +x2

2+

x3

6+ · · · =

∞∑k=0

xk

k!.

x

(1− x)2= x + 2x2 + 3x3 + 4x4 + · · · =

∞∑k=0

kxk ( for |x | < 1)

cos(x) = 1− x2

2+

x4

24+ · · · =

∞∑k=0

(−1)kx2k

(2k)!

arctan(x) = x − x3

3+

x5

5− x7

7+ · · · =

∞∑k=0

(−1)kx2k+1

2k + 1

I For any reasonable function f (x), there are coefficients ak such that

f (x) =∞∑k=0

akxk

(when x is sufficiently small). This is the Taylor series for f (x).

Taylor series

ex = exp(x) = 1 + x +x2

2+

x3

6+ · · · =

∞∑k=0

xk

k!.

x

(1− x)2= x + 2x2 + 3x3 + 4x4 + · · · =

∞∑k=0

kxk ( for |x | < 1)

cos(x) = 1− x2

2+

x4

24+ · · · =

∞∑k=0

(−1)kx2k

(2k)!


3+

x5

5− x7

7+ · · · =

∞∑k=0

(−1)kx2k+1

2k + 1


f (x) =∞∑k=0

akxk


Taylor series

ex = exp(x) = 1 + x +x2

2+

x3

6+ · · · =

∞∑k=0

xk

k!.

x

(1− x)2= x + 2x2 + 3x3 + 4x4 + · · · =

∞∑k=0

kxk ( for |x | < 1)

cos(x) = 1− x2

2+

x4

24+ · · · =

∞∑k=0

(−1)kx2k

(2k)!


3+

x5

5− x7

7+ · · · =

∞∑k=0

(−1)kx2k+1

2k + 1


f (x) =∞∑k=0

akxk


Taylor series

ex = exp(x) = 1 + x +x2

2+

x3

6+ · · · =

∞∑k=0

xk

k!.

x

(1− x)2= x + 2x2 + 3x3 + 4x4 + · · · =

∞∑k=0

kxk ( for |x | < 1)

cos(x) = 1− x2

2+

x4

24+ · · · =

∞∑k=0

(−1)kx2k

(2k)!


3+

x5

5− x7

7+ · · · =

∞∑k=0

(−1)kx2k+1

2k + 1


f (x) =∞∑k=0

akxk


Taylor series

ex = exp(x) = 1 + x +x2

2+

x3

6+ · · · =

∞∑k=0

xk

k!.

x

(1− x)2= x + 2x2 + 3x3 + 4x4 + · · · =

∞∑k=0

kxk ( for |x | < 1)

cos(x) = 1− x2

2+

x4

24+ · · · =

∞∑k=0

(−1)kx2k

(2k)!


3+

x5

5− x7

7+ · · · =

∞∑k=0

(−1)kx2k+1

2k + 1


f (x) =∞∑k=0

akxk


Taylor series

ex = exp(x) = 1 + x +x2

2+

x3

6+ · · · =

∞∑k=0

xk

k!.

x

(1− x)2= x + 2x2 + 3x3 + 4x4 + · · · =

∞∑k=0

kxk ( for |x | < 1)

cos(x) = 1− x2

2+

x4

24+ · · · =

∞∑k=0

(−1)kx2k

(2k)!


3+

x5

5− x7

7+ · · · =

∞∑k=0

(−1)kx2k+1

2k + 1


f (x) =∞∑k=0

akxk


Exceptions

Not every function has a Taylor series.

I f0(x) = 1/x does not, because f0(0) is undefined.

I f1(x) = |x | and f2(x) = x1/3 do not, because the slopes f ′1 (0) and f ′2 (0) arenot defined.

I f3(x) = ln(x) does not, because f ′3 (x) is undefined for x < 0.

I f4(x) = e−1/x2

does not, for a more subtle reason.

For a full explanation, see Level 3 complex analysis.

Exceptions





I f4(x) = e−1/x2



Exceptions





I f4(x) = e−1/x2



Exceptions





I f4(x) = e−1/x2



Exceptions





I f4(x) = e−1/x2



Exceptions





I f4(x) = e−1/x2



Exceptions





I f4(x) = e−1/x2



Truncated series

Often we only calculate with finitely many terms of the Taylor series.

tan(x) = x + x3/3 + 2x5/15 + O(x7)

The notation O(x7) means that there are extra terms involving powers xk

with k ≥ 7. The above is the 7th order Taylor series for tan(x). It is a goodapproximation to tan(x) if x is sufficiently small.

Truncated series


tan(x) = x + x3/3 + 2x5/15 + O(x7)



Truncated series


tan(x) = x + x3/3 + 2x5/15 + O(x7)



Truncated series


tan(x) = x + x3/3 + 2x5/15 + O(x7)


with k ≥ 7.

The above is the 7th order Taylor series for tan(x). It is a goodapproximation to tan(x) if x is sufficiently small.

Truncated series


tan(x) = x + x3/3 + 2x5/15 + O(x7)



Truncated series


tan(x) = x + x3/3 + 2x5/15 + O(x7)



tan(x) = x + O(x3)

Truncated series


tan(x) = x + x3/3 + 2x5/15 + O(x7)



tan(x) = x + x3/3 + O(x5)

Truncated series


tan(x) = x + x3/3 + 2x5/15 + O(x7)



tan(x) = x + x3/3 + 2x5/15 + O(x7)

Truncated series


tan(x) = x + x3/3 + 2x5/15 + O(x7)



tan(x) = x + x3/3 + 2x5/15 + 17x7/315 + O(x9)

Finding coefficients

y =∞∑k=0

akxk , where ak =

1

k!

dky

dxk

∣∣∣∣x=0

f (x) =∞∑k=0

akxk , where ak = f (k)(0)/k!

Example:

exp(k)(x) = · · · = exp′′′(x) = exp′′(x) = exp′(x) = exp(x) = ex

exp(k)(0) = · · · = exp′′′(0) = exp′′(0) = exp′(0) = exp(0) = 1

Thus ak = 1/k!, and exp(x) =∑

k xk/k!.


y =∞∑k=0

akxk , where ak =

1

k!

dky

dxk

∣∣∣∣x=0

f (x) =∞∑k=0


Example:




k xk/k!.


y =∞∑k=0

akxk , where ak =

1

k!

dky

dxk

∣∣∣∣x=0

f (x) =∞∑k=0


Example:




k xk/k!.


y =∞∑k=0

akxk , where ak =

1

k!

dky

dxk

∣∣∣∣x=0

f (x) =∞∑k=0


Example:




k xk/k!.


y =∞∑k=0

akxk , where ak =

1

k!

dky

dxk

∣∣∣∣x=0

f (x) =∞∑k=0


Example:




k xk/k!.


y =∞∑k=0

akxk , where ak =

1

k!

dky

dxk

∣∣∣∣x=0

f (x) =∞∑k=0


Example:




k xk/k!.

Another example

Take f (x) = sin(x).

f (x) = sin(x) f ′(x) = cos(x) f ′′(x) = − sin(x) f ′′′(x) = − cos(x)

f (4)(x) = sin(x) f (5)(x) = cos(x) f (6)(x) = − sin(x) f (7)(x) = − cos(x)


f (0) = 0 f ′(0) = 1 f ′′(0) = 0 f ′′′(0) = −1

f (4)(0) = 0 f (5)(0) = 1 f (6)(0) = 0 f (7)(0) = −1

f (8)(0) = 0 f (9)(0) = 1 f (10)(0) = 0 f (11)(0) = −1

a0 = 0 a1 = 1 a2 = 0 a3 = −1/3!

a4 = 0 a5 = 1/5! a6 = 0 a7 = −1/7!

a8 = 0 a9 = 1/9! a10 = 0 a11 = −1/11!

sin(x) = x − x3

3!+

x5

5!− x7

7!+

x9

9!− x11

11!+ . . . =

∞∑k=0

(−1)kx2k+1

(2k + 1)!

Another example





f (0) = 0 f ′(0) = 1 f ′′(0) = 0 f ′′′(0) = −1

f (4)(0) = 0 f (5)(0) = 1 f (6)(0) = 0 f (7)(0) = −1

f (8)(0) = 0 f (9)(0) = 1 f (10)(0) = 0 f (11)(0) = −1

a0 = 0 a1 = 1 a2 = 0 a3 = −1/3!

a4 = 0 a5 = 1/5! a6 = 0 a7 = −1/7!

a8 = 0 a9 = 1/9! a10 = 0 a11 = −1/11!

sin(x) = x − x3

3!+

x5

5!− x7

7!+

x9

9!− x11

11!+ . . . =

∞∑k=0

(−1)kx2k+1

(2k + 1)!

Another example





f (0) = 0 f ′(0) = 1 f ′′(0) = 0 f ′′′(0) = −1

f (4)(0) = 0 f (5)(0) = 1 f (6)(0) = 0 f (7)(0) = −1

f (8)(0) = 0 f (9)(0) = 1 f (10)(0) = 0 f (11)(0) = −1

a0 = 0 a1 = 1 a2 = 0 a3 = −1/3!

a4 = 0 a5 = 1/5! a6 = 0 a7 = −1/7!

a8 = 0 a9 = 1/9! a10 = 0 a11 = −1/11!

sin(x) = x − x3

3!+

x5

5!− x7

7!+

x9

9!− x11

11!+ . . . =

∞∑k=0

(−1)kx2k+1

(2k + 1)!

Another example





f (0) = 0 f ′(0) = 1 f ′′(0) = 0 f ′′′(0) = −1

f (4)(0) = 0 f (5)(0) = 1 f (6)(0) = 0 f (7)(0) = −1

f (8)(0) = 0 f (9)(0) = 1 f (10)(0) = 0 f (11)(0) = −1

a0 = 0 a1 = 1 a2 = 0 a3 = −1/3!

a4 = 0 a5 = 1/5! a6 = 0 a7 = −1/7!

a8 = 0 a9 = 1/9! a10 = 0 a11 = −1/11!

sin(x) = x − x3

3!+

x5

5!− x7

7!+

x9

9!− x11

11!+ . . . =

∞∑k=0

(−1)kx2k+1

(2k + 1)!

Another example





f (0) = 0 f ′(0) = 1 f ′′(0) = 0 f ′′′(0) = −1

f (4)(0) = 0 f (5)(0) = 1 f (6)(0) = 0 f (7)(0) = −1

f (8)(0) = 0 f (9)(0) = 1 f (10)(0) = 0 f (11)(0) = −1

a0 = 0 a1 = 1 a2 = 0 a3 = −1/3!

a4 = 0 a5 = 1/5! a6 = 0 a7 = −1/7!

a8 = 0 a9 = 1/9! a10 = 0 a11 = −1/11!

sin(x) = x − x3

3!+

x5

5!− x7

7!+

x9

9!− x11

11!+ . . . =

∞∑k=0

(−1)kx2k+1

(2k + 1)!

Another example





f (0) = 0 f ′(0) = 1 f ′′(0) = 0 f ′′′(0) = −1

f (4)(0) = 0 f (5)(0) = 1 f (6)(0) = 0 f (7)(0) = −1

f (8)(0) = 0 f (9)(0) = 1 f (10)(0) = 0 f (11)(0) = −1

a0 = 0 a1 = 1 a2 = 0 a3 = −1/3!

a4 = 0 a5 = 1/5! a6 = 0 a7 = −1/7!

a8 = 0 a9 = 1/9! a10 = 0 a11 = −1/11!

sin(x) = x − x3

3!+

x5

5!− x7

7!+

x9

9!− x11

11!+ . . . =

∞∑k=0

(−1)kx2k+1

(2k + 1)!

Another example





f (0) = 0 f ′(0) = 1 f ′′(0) = 0 f ′′′(0) = −1

f (4)(0) = 0 f (5)(0) = 1 f (6)(0) = 0 f (7)(0) = −1

f (8)(0) = 0 f (9)(0) = 1 f (10)(0) = 0 f (11)(0) = −1

a0 = 0 a1 = 1 a2 = 0 a3 = −1/3!

a4 = 0 a5 = 1/5! a6 = 0 a7 = −1/7!

a8 = 0 a9 = 1/9! a10 = 0 a11 = −1/11!

sin(x) = x − x3

3!+

x5

5!− x7

7!+

x9

9!− x11

11!+ . . . =

∞∑k=0

(−1)kx2k+1

(2k + 1)!

Another example





f (0) = 0 f ′(0) = 1 f ′′(0) = 0 f ′′′(0) = −1

f (4)(0) = 0 f (5)(0) = 1 f (6)(0) = 0 f (7)(0) = −1

f (8)(0) = 0 f (9)(0) = 1 f (10)(0) = 0 f (11)(0) = −1

a0 = 0 a1 = 1 a2 = 0 a3 = −1/3!

a4 = 0 a5 = 1/5! a6 = 0 a7 = −1/7!

a8 = 0 a9 = 1/9! a10 = 0 a11 = −1/11!

sin(x) = x − x3

3!+

x5

5!− x7

7!+

x9

9!− x11

11!+ . . . =

∞∑k=0

(−1)kx2k+1

(2k + 1)!

Another example





f (0) = 0 f ′(0) = 1 f ′′(0) = 0 f ′′′(0) = −1

f (4)(0) = 0 f (5)(0) = 1 f (6)(0) = 0 f (7)(0) = −1

f (8)(0) = 0 f (9)(0) = 1 f (10)(0) = 0 f (11)(0) = −1

a0 = 0 a1 = 1 a2 = 0 a3 = −1/3!

a4 = 0 a5 = 1/5! a6 = 0 a7 = −1/7!

a8 = 0 a9 = 1/9! a10 = 0 a11 = −1/11!

sin(x) = x − x3

3!+

x5

5!− x7

7!+

x9

9!− x11

11!+ . . . =

∞∑k=0

(−1)kx2k+1

(2k + 1)!

Another example





f (0) = 0 f ′(0) = 1 f ′′(0) = 0 f ′′′(0) = −1

f (4)(0) = 0 f (5)(0) = 1 f (6)(0) = 0 f (7)(0) = −1

f (8)(0) = 0 f (9)(0) = 1 f (10)(0) = 0 f (11)(0) = −1

a0 = 0 a1 = 1 a2 = 0 a3 = −1/3!

a4 = 0 a5 = 1/5! a6 = 0 a7 = −1/7!

a8 = 0 a9 = 1/9! a10 = 0 a11 = −1/11!

sin(x) = x − x3

3!+

x5

5!− x7

7!+

x9

9!− x11

11!+ . . . =

∞∑k=0

(−1)kx2k+1

(2k + 1)!

Another example





f (0) = 0 f ′(0) = 1 f ′′(0) = 0 f ′′′(0) = −1

f (4)(0) = 0 f (5)(0) = 1 f (6)(0) = 0 f (7)(0) = −1

f (8)(0) = 0 f (9)(0) = 1 f (10)(0) = 0 f (11)(0) = −1

a0 = 0 a1 = 1 a2 = 0 a3 = −1/3!

a4 = 0 a5 = 1/5! a6 = 0 a7 = −1/7!

a8 = 0 a9 = 1/9! a10 = 0 a11 = −1/11!

sin(x) = x − x3

3!+

x5

5!− x7

7!+

x9

9!− x11

11!+ . . . =

∞∑k=0

(−1)kx2k+1

(2k + 1)!

Another example





f (0) = 0 f ′(0) = 1 f ′′(0) = 0 f ′′′(0) = −1

f (4)(0) = 0 f (5)(0) = 1 f (6)(0) = 0 f (7)(0) = −1

f (8)(0) = 0 f (9)(0) = 1 f (10)(0) = 0 f (11)(0) = −1

a0 = 0 a1 = 1 a2 = 0 a3 = −1/3!

a4 = 0 a5 = 1/5! a6 = 0 a7 = −1/7!

a8 = 0 a9 = 1/9! a10 = 0 a11 = −1/11!

sin(x) = x − x3

3!+

x5

5!− x7

7!+

x9

9!− x11

11!+ . . .

=∞∑k=0

(−1)kx2k+1

(2k + 1)!

Another example





f (0) = 0 f ′(0) = 1 f ′′(0) = 0 f ′′′(0) = −1

f (4)(0) = 0 f (5)(0) = 1 f (6)(0) = 0 f (7)(0) = −1

f (8)(0) = 0 f (9)(0) = 1 f (10)(0) = 0 f (11)(0) = −1

a0 = 0 a1 = 1 a2 = 0 a3 = −1/3!

a4 = 0 a5 = 1/5! a6 = 0 a7 = −1/7!

a8 = 0 a9 = 1/9! a10 = 0 a11 = −1/11!

sin(x) = x − x3

3!+

x5

5!− x7

7!+

x9

9!− x11

11!+ . . . =

∞∑k=0

(−1)kx2k+1

(2k + 1)!

Other methods

It is often easiest to deduce a Taylor series from known series for otherfunctions.

e−x2=∑k

(−x2)k

k!=∑k

(−1)kx2k

k!

cosh(x) = (ex + e−x )/2 =∑k

xk + (−x)k

2 (k!)=∑keven

xk

k!=∑j

x2j

(2j)!

sinh(x)/x = (ex − e−x )/(2x) =∑k

xk − (−x)k

2x (k!)=∑kodd

xk−1

k!=∑j

x2j

(2j + 1)!

1/(1 − x) = 1 + x + x2 + x3 + . . . =∑k

xk

xd

dx

(1

1 − x

)= x

d

dx

∑k

xk = x∑k

k xk−1 =∑k

k xk

x/(1 − x)2 =∑k

k xk .

Other methods


e−x2=∑k

(−x2)k

k!=∑k

(−1)kx2k

k!

cosh(x) = (ex + e−x )/2 =∑k

xk + (−x)k

2 (k!)=∑keven

xk

k!=∑j

x2j

(2j)!

sinh(x)/x = (ex − e−x )/(2x) =∑k

xk − (−x)k

2x (k!)=∑kodd

xk−1

k!=∑j

x2j

(2j + 1)!

1/(1 − x) = 1 + x + x2 + x3 + . . . =∑k

xk

xd

dx

(1

1 − x

)= x

d

dx

∑k

xk = x∑k

k xk−1 =∑k

k xk

x/(1 − x)2 =∑k

k xk .

Other methods


e−x2

=∑k

(−x2)k

k!=∑k

(−1)kx2k

k!

cosh(x) = (ex + e−x )/2 =∑k

xk + (−x)k

2 (k!)=∑keven

xk

k!=∑j

x2j

(2j)!

sinh(x)/x = (ex − e−x )/(2x) =∑k

xk − (−x)k

2x (k!)=∑kodd

xk−1

k!=∑j

x2j

(2j + 1)!

1/(1 − x) = 1 + x + x2 + x3 + . . . =∑k

xk

xd

dx

(1

1 − x

)= x

d

dx

∑k

xk = x∑k

k xk−1 =∑k

k xk

x/(1 − x)2 =∑k

k xk .

Other methods


e−x2=∑k

(−x2)k

k!

=∑k

(−1)kx2k

k!

cosh(x) = (ex + e−x )/2 =∑k

xk + (−x)k

2 (k!)=∑keven

xk

k!=∑j

x2j

(2j)!

sinh(x)/x = (ex − e−x )/(2x) =∑k

xk − (−x)k

2x (k!)=∑kodd

xk−1

k!=∑j

x2j

(2j + 1)!

1/(1 − x) = 1 + x + x2 + x3 + . . . =∑k

xk

xd

dx

(1

1 − x

)= x

d

dx

∑k

xk = x∑k

k xk−1 =∑k

k xk

x/(1 − x)2 =∑k

k xk .

Other methods


e−x2=∑k

(−x2)k

k!=∑k

(−1)kx2k

k!

cosh(x) = (ex + e−x )/2 =∑k

xk + (−x)k

2 (k!)=∑keven

xk

k!=∑j

x2j

(2j)!

sinh(x)/x = (ex − e−x )/(2x) =∑k

xk − (−x)k

2x (k!)=∑kodd

xk−1

k!=∑j

x2j

(2j + 1)!

1/(1 − x) = 1 + x + x2 + x3 + . . . =∑k

xk

xd

dx

(1

1 − x

)= x

d

dx

∑k

xk = x∑k

k xk−1 =∑k

k xk

x/(1 − x)2 =∑k

k xk .

Other methods


e−x2=∑k

(−x2)k

k!=∑k

(−1)kx2k

k!

cosh(x)

= (ex + e−x )/2 =∑k

xk + (−x)k

2 (k!)=∑keven

xk

k!=∑j

x2j

(2j)!

sinh(x)/x = (ex − e−x )/(2x) =∑k

xk − (−x)k

2x (k!)=∑kodd

xk−1

k!=∑j

x2j

(2j + 1)!

1/(1 − x) = 1 + x + x2 + x3 + . . . =∑k

xk

xd

dx

(1

1 − x

)= x

d

dx

∑k

xk = x∑k

k xk−1 =∑k

k xk

x/(1 − x)2 =∑k

k xk .

Other methods


e−x2=∑k

(−x2)k

k!=∑k

(−1)kx2k

k!

cosh(x) = (ex + e−x )/2

=∑k

xk + (−x)k

2 (k!)=∑keven

xk

k!=∑j

x2j

(2j)!

sinh(x)/x = (ex − e−x )/(2x) =∑k

xk − (−x)k

2x (k!)=∑kodd

xk−1

k!=∑j

x2j

(2j + 1)!

1/(1 − x) = 1 + x + x2 + x3 + . . . =∑k

xk

xd

dx

(1

1 − x

)= x

d

dx

∑k

xk = x∑k

k xk−1 =∑k

k xk

x/(1 − x)2 =∑k

k xk .

Other methods


e−x2=∑k

(−x2)k

k!=∑k

(−1)kx2k

k!

cosh(x) = (ex + e−x )/2 =∑k

xk + (−x)k

2 (k!)

=∑keven

xk

k!=∑j

x2j

(2j)!

sinh(x)/x = (ex − e−x )/(2x) =∑k

xk − (−x)k

2x (k!)=∑kodd

xk−1

k!=∑j

x2j

(2j + 1)!

1/(1 − x) = 1 + x + x2 + x3 + . . . =∑k

xk

xd

dx

(1

1 − x

)= x

d

dx

∑k

xk = x∑k

k xk−1 =∑k

k xk

x/(1 − x)2 =∑k

k xk .

Other methods


e−x2=∑k

(−x2)k

k!=∑k

(−1)kx2k

k!

cosh(x) = (ex + e−x )/2 =∑k

xk + (−x)k

2 (k!)=∑keven

xk

k!

=∑j

x2j

(2j)!

sinh(x)/x = (ex − e−x )/(2x) =∑k

xk − (−x)k

2x (k!)=∑kodd

xk−1

k!=∑j

x2j

(2j + 1)!

1/(1 − x) = 1 + x + x2 + x3 + . . . =∑k

xk

xd

dx

(1

1 − x

)= x

d

dx

∑k

xk = x∑k

k xk−1 =∑k

k xk

x/(1 − x)2 =∑k

k xk .

Other methods


e−x2=∑k

(−x2)k

k!=∑k

(−1)kx2k

k!

cosh(x) = (ex + e−x )/2 =∑k

xk + (−x)k

2 (k!)=∑keven

xk

k!=∑j

x2j

(2j)!

sinh(x)/x = (ex − e−x )/(2x) =∑k

xk − (−x)k

2x (k!)=∑kodd

xk−1

k!=∑j

x2j

(2j + 1)!

1/(1 − x) = 1 + x + x2 + x3 + . . . =∑k

xk

xd

dx

(1

1 − x

)= x

d

dx

∑k

xk = x∑k

k xk−1 =∑k

k xk

x/(1 − x)2 =∑k

k xk .

Other methods


e−x2=∑k

(−x2)k

k!=∑k

(−1)kx2k

k!

cosh(x) = (ex + e−x )/2 =∑k

xk + (−x)k

2 (k!)=∑keven

xk

k!=∑j

x2j

(2j)!

sinh(x)/x

= (ex − e−x )/(2x) =∑k

xk − (−x)k

2x (k!)=∑kodd

xk−1

k!=∑j

x2j

(2j + 1)!

1/(1 − x) = 1 + x + x2 + x3 + . . . =∑k

xk

xd

dx

(1

1 − x

)= x

d

dx

∑k

xk = x∑k

k xk−1 =∑k

k xk

x/(1 − x)2 =∑k

k xk .

Other methods


e−x2=∑k

(−x2)k

k!=∑k

(−1)kx2k

k!

cosh(x) = (ex + e−x )/2 =∑k

xk + (−x)k

2 (k!)=∑keven

xk

k!=∑j

x2j

(2j)!

sinh(x)/x = (ex − e−x )/(2x)

=∑k

xk − (−x)k

2x (k!)=∑kodd

xk−1

k!=∑j

x2j

(2j + 1)!

1/(1 − x) = 1 + x + x2 + x3 + . . . =∑k

xk

xd

dx

(1

1 − x

)= x

d

dx

∑k

xk = x∑k

k xk−1 =∑k

k xk

x/(1 − x)2 =∑k

k xk .

Other methods


e−x2=∑k

(−x2)k

k!=∑k

(−1)kx2k

k!

cosh(x) = (ex + e−x )/2 =∑k

xk + (−x)k

2 (k!)=∑keven

xk

k!=∑j

x2j

(2j)!

sinh(x)/x = (ex − e−x )/(2x) =∑k

xk − (−x)k

2x (k!)

=∑kodd

xk−1

k!=∑j

x2j

(2j + 1)!

1/(1 − x) = 1 + x + x2 + x3 + . . . =∑k

xk

xd

dx

(1

1 − x

)= x

d

dx

∑k

xk = x∑k

k xk−1 =∑k

k xk

x/(1 − x)2 =∑k

k xk .

Other methods


e−x2=∑k

(−x2)k

k!=∑k

(−1)kx2k

k!

cosh(x) = (ex + e−x )/2 =∑k

xk + (−x)k

2 (k!)=∑keven

xk

k!=∑j

x2j

(2j)!

sinh(x)/x = (ex − e−x )/(2x) =∑k

xk − (−x)k

2x (k!)=∑kodd

xk−1

k!

=∑j

x2j

(2j + 1)!

1/(1 − x) = 1 + x + x2 + x3 + . . . =∑k

xk

xd

dx

(1

1 − x

)= x

d

dx

∑k

xk = x∑k

k xk−1 =∑k

k xk

x/(1 − x)2 =∑k

k xk .

Other methods


e−x2=∑k

(−x2)k

k!=∑k

(−1)kx2k

k!

cosh(x) = (ex + e−x )/2 =∑k

xk + (−x)k

2 (k!)=∑keven

xk

k!=∑j

x2j

(2j)!

sinh(x)/x = (ex − e−x )/(2x) =∑k

xk − (−x)k

2x (k!)=∑kodd

xk−1

k!=∑j

x2j

(2j + 1)!

1/(1 − x) = 1 + x + x2 + x3 + . . . =∑k

xk

xd

dx

(1

1 − x

)= x

d

dx

∑k

xk = x∑k

k xk−1 =∑k

k xk

x/(1 − x)2 =∑k

k xk .

Other methods


e−x2=∑k

(−x2)k

k!=∑k

(−1)kx2k

k!

cosh(x) = (ex + e−x )/2 =∑k

xk + (−x)k

2 (k!)=∑keven

xk

k!=∑j

x2j

(2j)!

sinh(x)/x = (ex − e−x )/(2x) =∑k

xk − (−x)k

2x (k!)=∑kodd

xk−1

k!=∑j

x2j

(2j + 1)!

1/(1 − x)

= 1 + x + x2 + x3 + . . . =∑k

xk

xd

dx

(1

1 − x

)= x

d

dx

∑k

xk = x∑k

k xk−1 =∑k

k xk

x/(1 − x)2 =∑k

k xk .

Other methods


e−x2=∑k

(−x2)k

k!=∑k

(−1)kx2k

k!

cosh(x) = (ex + e−x )/2 =∑k

xk + (−x)k

2 (k!)=∑keven

xk

k!=∑j

x2j

(2j)!

sinh(x)/x = (ex − e−x )/(2x) =∑k

xk − (−x)k

2x (k!)=∑kodd

xk−1

k!=∑j

x2j

(2j + 1)!

1/(1 − x) = 1 + x + x2 + x3 + . . .

=∑k

xk

xd

dx

(1

1 − x

)= x

d

dx

∑k

xk = x∑k

k xk−1 =∑k

k xk

x/(1 − x)2 =∑k

k xk .

Other methods


e−x2=∑k

(−x2)k

k!=∑k

(−1)kx2k

k!

cosh(x) = (ex + e−x )/2 =∑k

xk + (−x)k

2 (k!)=∑keven

xk

k!=∑j

x2j

(2j)!

sinh(x)/x = (ex − e−x )/(2x) =∑k

xk − (−x)k

2x (k!)=∑kodd

xk−1

k!=∑j

x2j

(2j + 1)!

1/(1 − x) = 1 + x + x2 + x3 + . . . =∑k

xk

xd

dx

(1

1 − x

)= x

d

dx

∑k

xk = x∑k

k xk−1 =∑k

k xk

x/(1 − x)2 =∑k

k xk .

Other methods


e−x2=∑k

(−x2)k

k!=∑k

(−1)kx2k

k!

cosh(x) = (ex + e−x )/2 =∑k

xk + (−x)k

2 (k!)=∑keven

xk

k!=∑j

x2j

(2j)!

sinh(x)/x = (ex − e−x )/(2x) =∑k

xk − (−x)k

2x (k!)=∑kodd

xk−1

k!=∑j

x2j

(2j + 1)!

1/(1 − x) = 1 + x + x2 + x3 + . . . =∑k

xk

xd

dx

(1

1 − x

)

= xd

dx

∑k

xk = x∑k

k xk−1 =∑k

k xk

x/(1 − x)2 =∑k

k xk .

Other methods


e−x2=∑k

(−x2)k

k!=∑k

(−1)kx2k

k!

cosh(x) = (ex + e−x )/2 =∑k

xk + (−x)k

2 (k!)=∑keven

xk

k!=∑j

x2j

(2j)!

sinh(x)/x = (ex − e−x )/(2x) =∑k

xk − (−x)k

2x (k!)=∑kodd

xk−1

k!=∑j

x2j

(2j + 1)!

1/(1 − x) = 1 + x + x2 + x3 + . . . =∑k

xk

xd

dx

(1

1 − x

)= x

d

dx

∑k

xk

= x∑k

k xk−1 =∑k

k xk

x/(1 − x)2 =∑k

k xk .

Other methods


e−x2=∑k

(−x2)k

k!=∑k

(−1)kx2k

k!

cosh(x) = (ex + e−x )/2 =∑k

xk + (−x)k

2 (k!)=∑keven

xk

k!=∑j

x2j

(2j)!

sinh(x)/x = (ex − e−x )/(2x) =∑k

xk − (−x)k

2x (k!)=∑kodd

xk−1

k!=∑j

x2j

(2j + 1)!

1/(1 − x) = 1 + x + x2 + x3 + . . . =∑k

xk

xd

dx

(1

1 − x

)= x

d

dx

∑k

xk = x∑k

k xk−1

=∑k

k xk

x/(1 − x)2 =∑k

k xk .

Other methods


e−x2=∑k

(−x2)k

k!=∑k

(−1)kx2k

k!

cosh(x) = (ex + e−x )/2 =∑k

xk + (−x)k

2 (k!)=∑keven

xk

k!=∑j

x2j

(2j)!

sinh(x)/x = (ex − e−x )/(2x) =∑k

xk − (−x)k

2x (k!)=∑kodd

xk−1

k!=∑j

x2j

(2j + 1)!

1/(1 − x) = 1 + x + x2 + x3 + . . . =∑k

xk

xd

dx

(1

1 − x

)= x

d

dx

∑k

xk = x∑k

k xk−1 =∑k

k xk

x/(1 − x)2 =∑k

k xk .

Other methods


e−x2=∑k

(−x2)k

k!=∑k

(−1)kx2k

k!

cosh(x) = (ex + e−x )/2 =∑k

xk + (−x)k

2 (k!)=∑keven

xk

k!=∑j

x2j

(2j)!

sinh(x)/x = (ex − e−x )/(2x) =∑k

xk − (−x)k

2x (k!)=∑kodd

xk−1

k!=∑j

x2j

(2j + 1)!

1/(1 − x) = 1 + x + x2 + x3 + . . . =∑k

xk

xd

dx

(1

1 − x

)= x

d

dx

∑k

xk = x∑k

k xk−1 =∑k

k xk

x/(1 − x)2

=∑k

k xk .

Other methods


e−x2=∑k

(−x2)k

k!=∑k

(−1)kx2k

k!

cosh(x) = (ex + e−x )/2 =∑k

xk + (−x)k

2 (k!)=∑keven

xk

k!=∑j

x2j

(2j)!

sinh(x)/x = (ex − e−x )/(2x) =∑k

xk − (−x)k

2x (k!)=∑kodd

xk−1

k!=∑j

x2j

(2j + 1)!

1/(1 − x) = 1 + x + x2 + x3 + . . . =∑k

xk

xd

dx

(1

1 − x

)= x

d

dx

∑k

xk = x∑k

k xk−1 =∑k

k xk

x/(1 − x)2 =∑k

k xk .

Odd and even functions

Recall that f (x) is even if f (−x) = f (x), and odd if f (−x) = −f (x).For example, cos(x) is even and sin(x) is odd. If

f (x) =∑k

akxk =

∑k even

akxk +

∑k odd

akxk

thenf (−x) =

∑k

ak(−x)k =∑

k even

akxk −

∑k odd

akxk .

Thus f (x) is even iff the Taylor series involves only even powers of x , andf (x) is odd iff the Taylor series involves only odd powers of x .

sin(x) = x − x3

3!+

x5

5!− x7

7!+

x9

9!− x11

11!+ . . . =

∞∑k=0

(−1)kx2k+1

(2k + 1)!

cos(x) = 1− x2

2!+

x4

4!− x6

6!+

x8

8!− x10

10!+ . . . =

∞∑k=0

(−1)kx2k

(2k)!


Recall that f (x) is even if f (−x) = f (x), and odd if f (−x) = −f (x).

For example, cos(x) is even and sin(x) is odd. If

f (x) =∑k

akxk =

∑k even

akxk +

∑k odd

akxk

thenf (−x) =

∑k

ak(−x)k =∑

k even

akxk −

∑k odd

akxk .


sin(x) = x − x3

3!+

x5

5!− x7

7!+

x9

9!− x11

11!+ . . . =

∞∑k=0

(−1)kx2k+1

(2k + 1)!

cos(x) = 1− x2

2!+

x4

4!− x6

6!+

x8

8!− x10

10!+ . . . =

∞∑k=0

(−1)kx2k

(2k)!


Recall that f (x) is even if f (−x) = f (x), and odd if f (−x) = −f (x).For example, cos(x) is even and sin(x) is odd.

If

f (x) =∑k

akxk =

∑k even

akxk +

∑k odd

akxk

thenf (−x) =

∑k

ak(−x)k =∑

k even

akxk −

∑k odd

akxk .


sin(x) = x − x3

3!+

x5

5!− x7

7!+

x9

9!− x11

11!+ . . . =

∞∑k=0

(−1)kx2k+1

(2k + 1)!

cos(x) = 1− x2

2!+

x4

4!− x6

6!+

x8

8!− x10

10!+ . . . =

∞∑k=0

(−1)kx2k

(2k)!



f (x) =∑k

akxk =

∑k even

akxk +

∑k odd

akxk

thenf (−x) =

∑k

ak(−x)k =∑

k even

akxk −

∑k odd

akxk .


sin(x) = x − x3

3!+

x5

5!− x7

7!+

x9

9!− x11

11!+ . . . =

∞∑k=0

(−1)kx2k+1

(2k + 1)!

cos(x) = 1− x2

2!+

x4

4!− x6

6!+

x8

8!− x10

10!+ . . . =

∞∑k=0

(−1)kx2k

(2k)!



f (x) =∑k

akxk =

∑k even

akxk +

∑k odd

akxk

thenf (−x) =

∑k

ak(−x)k =∑

k even

akxk −

∑k odd

akxk .


sin(x) = x − x3

3!+

x5

5!− x7

7!+

x9

9!− x11

11!+ . . . =

∞∑k=0

(−1)kx2k+1

(2k + 1)!

cos(x) = 1− x2

2!+

x4

4!− x6

6!+

x8

8!− x10

10!+ . . . =

∞∑k=0

(−1)kx2k

(2k)!



f (x) =∑k

akxk =

∑k even

akxk +

∑k odd

akxk

thenf (−x) =

∑k

ak(−x)k =∑

k even

akxk −

∑k odd

akxk .


sin(x) = x − x3

3!+

x5

5!− x7

7!+

x9

9!− x11

11!+ . . . =

∞∑k=0

(−1)kx2k+1

(2k + 1)!

cos(x) = 1− x2

2!+

x4

4!− x6

6!+

x8

8!− x10

10!+ . . . =

∞∑k=0

(−1)kx2k

(2k)!

Algebra of series

tan(x) = x + 13x3 + 2

15x5 + O(x7)

tan(x)2 = (x + 13x3 + 2

15x5)2 + O(x7)

= x2 + 13x4 + 2

15x6 +

13x4 + 1

9x6 + 2

45x8

215x6 + 2

45x8 + 4

225x10 + O(x7)

= x2 + 23x4 + 17

45x6 + O(x7).

Algebra of series

tan(x) = x + 13x3 + 2

15x5 + O(x7)

tan(x)2 = (x + 13x3 + 2

15x5)2 + O(x7)

= x2 + 13x4 + 2

15x6 +

13x4 + 1

9x6 + 2

45x8

215x6 + 2

45x8 + 4

225x10 + O(x7)

= x2 + 23x4 + 17

45x6 + O(x7).

Algebra of series

tan(x) = x + 13x3 + 2

15x5 + O(x7)

tan(x)2 = (x + 13x3 + 2

15x5)2 + O(x7)

= x2 + 13x4 + 2

15x6 +

13x4 + 1

9x6 + 2

45x8

215x6 + 2

45x8 + 4

225x10 + O(x7)

= x2 + 23x4 + 17

45x6 + O(x7).

Algebra of series

tan(x) = x + 13x3 + 2

15x5 + O(x7)

tan(x)2 = (x + 13x3 + 2

15x5)2 + O(x7)

= x2 + 13x4 + 2

15x6 +

13x4 + 1

9x6 + 2

45x8

215x6 + 2

45x8 + 4

225x10 + O(x7)

= x2 + 23x4 + 17

45x6 + O(x7).

Algebra of series

tan(x) = x + 13x3 + 2

15x5 + O(x7)

tan(x)2 = (x + 13x3 + 2

15x5)2 + O(x7)

= x2 + 13x4 + 2

15x6 +

13x4 + 1

9x6 + 2

45x8

215x6 + 2

45x8 + 4

225x10 + O(x7)

= x2 + 23x4 + 17

45x6 + O(x7).

Expansion about other points

We can also expand f (x) in terms of powers (x − α)k , for any α. Moreprecisely,

f (x) =∑∞

k=0 bk(x − α)k , where bk = f (k)(α)/k!

ln′(x) = x−1 ln′′(x) = −x−2 ln′′′(x) = 2x−3 ln(4)(x) = −6x−4

ln(1) = 0 ln′(1) = 1 ln′′(1) = −1 ln′′′(1) = 2 ln(4)(1) = −6

b0 = 0 b1 = 1 b2 = −1/2 b3 = 2/3! = 1/3 b4 = −6/4! = −1/4

ln(x) = (x − 1)− (x − 1)2/2 + (x − 1)3/3− (x − 1)4/4 + O((x − 1)5).


We can also expand f (x) in terms of powers (x − α)k , for any α.

Moreprecisely,

f (x) =∑∞



ln(1) = 0 ln′(1) = 1 ln′′(1) = −1 ln′′′(1) = 2 ln(4)(1) = −6

b0 = 0 b1 = 1 b2 = −1/2 b3 = 2/3! = 1/3 b4 = −6/4! = −1/4

ln(x) = (x − 1)− (x − 1)2/2 + (x − 1)3/3− (x − 1)4/4 + O((x − 1)5).



f (x) =∑∞



ln(1) = 0 ln′(1) = 1 ln′′(1) = −1 ln′′′(1) = 2 ln(4)(1) = −6

b0 = 0 b1 = 1 b2 = −1/2 b3 = 2/3! = 1/3 b4 = −6/4! = −1/4

ln(x) = (x − 1)− (x − 1)2/2 + (x − 1)3/3− (x − 1)4/4 + O((x − 1)5).



f (x) =∑∞


ln′(x) = x−1

ln′′(x) = −x−2 ln′′′(x) = 2x−3 ln(4)(x) = −6x−4

ln(1) = 0 ln′(1) = 1 ln′′(1) = −1 ln′′′(1) = 2 ln(4)(1) = −6

b0 = 0 b1 = 1 b2 = −1/2 b3 = 2/3! = 1/3 b4 = −6/4! = −1/4

ln(x) = (x − 1)− (x − 1)2/2 + (x − 1)3/3− (x − 1)4/4 + O((x − 1)5).



f (x) =∑∞


ln′(x) = x−1 ln′′(x) = −x−2

ln′′′(x) = 2x−3 ln(4)(x) = −6x−4

ln(1) = 0 ln′(1) = 1 ln′′(1) = −1 ln′′′(1) = 2 ln(4)(1) = −6

b0 = 0 b1 = 1 b2 = −1/2 b3 = 2/3! = 1/3 b4 = −6/4! = −1/4

ln(x) = (x − 1)− (x − 1)2/2 + (x − 1)3/3− (x − 1)4/4 + O((x − 1)5).



f (x) =∑∞


ln′(x) = x−1 ln′′(x) = −x−2 ln′′′(x) = 2x−3

ln(4)(x) = −6x−4

ln(1) = 0 ln′(1) = 1 ln′′(1) = −1 ln′′′(1) = 2 ln(4)(1) = −6

b0 = 0 b1 = 1 b2 = −1/2 b3 = 2/3! = 1/3 b4 = −6/4! = −1/4

ln(x) = (x − 1)− (x − 1)2/2 + (x − 1)3/3− (x − 1)4/4 + O((x − 1)5).



f (x) =∑∞



ln(1) = 0 ln′(1) = 1 ln′′(1) = −1 ln′′′(1) = 2 ln(4)(1) = −6

b0 = 0 b1 = 1 b2 = −1/2 b3 = 2/3! = 1/3 b4 = −6/4! = −1/4

ln(x) = (x − 1)− (x − 1)2/2 + (x − 1)3/3− (x − 1)4/4 + O((x − 1)5).



f (x) =∑∞



ln(1) = 0 ln′(1) = 1 ln′′(1) = −1 ln′′′(1) = 2 ln(4)(1) = −6

b0 = 0 b1 = 1 b2 = −1/2 b3 = 2/3! = 1/3 b4 = −6/4! = −1/4

ln(x) = (x − 1)− (x − 1)2/2 + (x − 1)3/3− (x − 1)4/4 + O((x − 1)5).



f (x) =∑∞



ln(1) = 0 ln′(1) = 1 ln′′(1) = −1 ln′′′(1) = 2 ln(4)(1) = −6

b0 = 0 b1 = 1 b2 = −1/2 b3 = 2/3! = 1/3 b4 = −6/4! = −1/4

ln(x) = (x − 1)− (x − 1)2/2 + (x − 1)3/3− (x − 1)4/4 + O((x − 1)5).



f (x) =∑∞



ln(1) = 0 ln′(1) = 1 ln′′(1) = −1 ln′′′(1) = 2 ln(4)(1) = −6

b0 = 0 b1 = 1 b2 = −1/2 b3 = 2/3! = 1/3 b4 = −6/4! = −1/4

ln(x) = (x − 1)− (x − 1)2/2 + (x − 1)3/3− (x − 1)4/4 + O((x − 1)5).

More examples

We will find the series for tan(x) near x = π4

.

f (x) = tan(x)

f ′(x) =1

cos(x)2

f ′′(x) = −2 cos(x)−3.− sin(x) =2 sin(x)

cos(x)3

f (π4

) = 1 f ′(π4

) =1

(2−1/2)2= 2 f ′′(π

4) =

2.2−1/2

(2−1/2)3= 4

a0 = 1/0! = 1 a1 = 2/1! = 2 a2 = 4/2! = 2

tan(x) = 1 + 2(x − π4

) + 2(x − π4

)2 + O((x − π4

)3).

More examples


.

f (x) = tan(x)

f ′(x) =1

cos(x)2

f ′′(x) = −2 cos(x)−3.− sin(x) =2 sin(x)

cos(x)3

f (π4

) = 1 f ′(π4

) =1

(2−1/2)2= 2 f ′′(π

4) =

2.2−1/2

(2−1/2)3= 4

a0 = 1/0! = 1 a1 = 2/1! = 2 a2 = 4/2! = 2

tan(x) = 1 + 2(x − π4

) + 2(x − π4

)2 + O((x − π4

)3).

More examples


.

f (x) = tan(x)

f ′(x) =1

cos(x)2

f ′′(x) = −2 cos(x)−3.− sin(x) =2 sin(x)

cos(x)3

f (π4

) = 1 f ′(π4

) =1

(2−1/2)2= 2 f ′′(π

4) =

2.2−1/2

(2−1/2)3= 4

a0 = 1/0! = 1 a1 = 2/1! = 2 a2 = 4/2! = 2

tan(x) = 1 + 2(x − π4

) + 2(x − π4

)2 + O((x − π4

)3).

More examples


.

f (x) = tan(x)

f ′(x) =1

cos(x)2

f ′′(x) = −2 cos(x)−3.− sin(x) =2 sin(x)

cos(x)3

f (π4

) = 1 f ′(π4

) =1

(2−1/2)2= 2 f ′′(π

4) =

2.2−1/2

(2−1/2)3= 4

a0 = 1/0! = 1 a1 = 2/1! = 2 a2 = 4/2! = 2

tan(x) = 1 + 2(x − π4

) + 2(x − π4

)2 + O((x − π4

)3).

More examples


.

f (x) = tan(x)

f ′(x) =1

cos(x)2

f ′′(x) = −2 cos(x)−3.− sin(x) =2 sin(x)

cos(x)3

f (π4

) = 1

f ′(π4

) =1

(2−1/2)2= 2 f ′′(π

4) =

2.2−1/2

(2−1/2)3= 4

a0 = 1/0! = 1 a1 = 2/1! = 2 a2 = 4/2! = 2

tan(x) = 1 + 2(x − π4

) + 2(x − π4

)2 + O((x − π4

)3).

More examples


.

f (x) = tan(x)

f ′(x) =1

cos(x)2

f ′′(x) = −2 cos(x)−3.− sin(x) =2 sin(x)

cos(x)3

f (π4

) = 1 f ′(π4

) =1

(2−1/2)2= 2

f ′′(π4

) =2.2−1/2

(2−1/2)3= 4

a0 = 1/0! = 1 a1 = 2/1! = 2 a2 = 4/2! = 2

tan(x) = 1 + 2(x − π4

) + 2(x − π4

)2 + O((x − π4

)3).

More examples


.

f (x) = tan(x)

f ′(x) =1

cos(x)2

f ′′(x) = −2 cos(x)−3.− sin(x) =2 sin(x)

cos(x)3

f (π4

) = 1 f ′(π4

) =1

(2−1/2)2= 2 f ′′(π

4) =

2.2−1/2

(2−1/2)3= 4

a0 = 1/0! = 1 a1 = 2/1! = 2 a2 = 4/2! = 2

tan(x) = 1 + 2(x − π4

) + 2(x − π4

)2 + O((x − π4

)3).

More examples


.

f (x) = tan(x)

f ′(x) =1

cos(x)2

f ′′(x) = −2 cos(x)−3.− sin(x) =2 sin(x)

cos(x)3

f (π4

) = 1 f ′(π4

) =1

(2−1/2)2= 2 f ′′(π

4) =

2.2−1/2

(2−1/2)3= 4

a0 = 1/0! = 1 a1 = 2/1! = 2 a2 = 4/2! = 2

tan(x) = 1 + 2(x − π4

) + 2(x − π4

)2 + O((x − π4

)3).

More examples


.

f (x) = tan(x)

f ′(x) =1

cos(x)2

f ′′(x) = −2 cos(x)−3.− sin(x) =2 sin(x)

cos(x)3

f (π4

) = 1 f ′(π4

) =1

(2−1/2)2= 2 f ′′(π

4) =

2.2−1/2

(2−1/2)3= 4

a0 = 1/0! = 1 a1 = 2/1! = 2 a2 = 4/2! = 2

tan(x) = 1 + 2(x − π4

) + 2(x − π4

)2 + O((x − π4

)3).

More examples

Consider y = x/(ex − 1).

ex = 1 + x + x2/2 + x3/6 + O(x4)

ex − 1 = x + x2/2 + x3/6 + O(x4)

1

y=

ex − 1

x= 1 + x/2 + x2/6 + O(x3) = 1 + u + O(x3) u = x/2 + x2/6

y =1

1 + u= 1− u + u2 + O(u3) = 1− u + u2 + O(x3)

u2 = x2/4 + x3/6 + x4/36 = x2/4 + O(x3)x

ex − 1= 1− u + u2 + O(x3)

= 1− x/2− x2/6 + x2/4 + O(x3) = 1− x/2 + x2/12 + O(x3)

More examples


ex = 1 + x + x2/2 + x3/6 + O(x4)

ex − 1 = x + x2/2 + x3/6 + O(x4)

1

y=

ex − 1

x= 1 + x/2 + x2/6 + O(x3) = 1 + u + O(x3) u = x/2 + x2/6

y =1

1 + u= 1− u + u2 + O(u3) = 1− u + u2 + O(x3)

u2 = x2/4 + x3/6 + x4/36 = x2/4 + O(x3)x

ex − 1= 1− u + u2 + O(x3)

= 1− x/2− x2/6 + x2/4 + O(x3) = 1− x/2 + x2/12 + O(x3)

More examples


ex = 1 + x + x2/2 + x3/6 + O(x4)

ex − 1 = x + x2/2 + x3/6 + O(x4)

1

y=

ex − 1

x= 1 + x/2 + x2/6 + O(x3) = 1 + u + O(x3) u = x/2 + x2/6

y =1

1 + u= 1− u + u2 + O(u3) = 1− u + u2 + O(x3)

u2 = x2/4 + x3/6 + x4/36 = x2/4 + O(x3)x

ex − 1= 1− u + u2 + O(x3)

= 1− x/2− x2/6 + x2/4 + O(x3) = 1− x/2 + x2/12 + O(x3)

More examples


ex = 1 + x + x2/2 + x3/6 + O(x4)

ex − 1 = x + x2/2 + x3/6 + O(x4)

1

y=

ex − 1

x= 1 + x/2 + x2/6 + O(x3)

= 1 + u + O(x3) u = x/2 + x2/6

y =1

1 + u= 1− u + u2 + O(u3) = 1− u + u2 + O(x3)

u2 = x2/4 + x3/6 + x4/36 = x2/4 + O(x3)x

ex − 1= 1− u + u2 + O(x3)

= 1− x/2− x2/6 + x2/4 + O(x3) = 1− x/2 + x2/12 + O(x3)

More examples


ex = 1 + x + x2/2 + x3/6 + O(x4)

ex − 1 = x + x2/2 + x3/6 + O(x4)

1

y=

ex − 1

x= 1 + x/2 + x2/6 + O(x3) = 1 + u + O(x3) u = x/2 + x2/6

y =1

1 + u= 1− u + u2 + O(u3) = 1− u + u2 + O(x3)

u2 = x2/4 + x3/6 + x4/36 = x2/4 + O(x3)x

ex − 1= 1− u + u2 + O(x3)

= 1− x/2− x2/6 + x2/4 + O(x3) = 1− x/2 + x2/12 + O(x3)

More examples


ex = 1 + x + x2/2 + x3/6 + O(x4)

ex − 1 = x + x2/2 + x3/6 + O(x4)

1

y=

ex − 1

x= 1 + x/2 + x2/6 + O(x3) = 1 + u + O(x3) u = x/2 + x2/6

y =1

1 + u= 1− u + u2 + O(u3) = 1− u + u2 + O(x3)

u2 = x2/4 + x3/6 + x4/36 = x2/4 + O(x3)x

ex − 1= 1− u + u2 + O(x3)

= 1− x/2− x2/6 + x2/4 + O(x3) = 1− x/2 + x2/12 + O(x3)

More examples


ex = 1 + x + x2/2 + x3/6 + O(x4)

ex − 1 = x + x2/2 + x3/6 + O(x4)

1

y=

ex − 1

x= 1 + x/2 + x2/6 + O(x3) = 1 + u + O(x3) u = x/2 + x2/6

y =1

1 + u= 1− u + u2 + O(u3) = 1− u + u2 + O(x3)

u2 = x2/4 + x3/6 + x4/36

= x2/4 + O(x3)x

ex − 1= 1− u + u2 + O(x3)

= 1− x/2− x2/6 + x2/4 + O(x3) = 1− x/2 + x2/12 + O(x3)

More examples


ex = 1 + x + x2/2 + x3/6 + O(x4)

ex − 1 = x + x2/2 + x3/6 + O(x4)

1

y=

ex − 1

x= 1 + x/2 + x2/6 + O(x3) = 1 + u + O(x3) u = x/2 + x2/6

y =1

1 + u= 1− u + u2 + O(u3) = 1− u + u2 + O(x3)

u2 = x2/4 + x3/6 + x4/36 = x2/4 + O(x3)

x

ex − 1= 1− u + u2 + O(x3)

= 1− x/2− x2/6 + x2/4 + O(x3) = 1− x/2 + x2/12 + O(x3)

More examples


ex = 1 + x + x2/2 + x3/6 + O(x4)

ex − 1 = x + x2/2 + x3/6 + O(x4)

1

y=

ex − 1

x= 1 + x/2 + x2/6 + O(x3) = 1 + u + O(x3) u = x/2 + x2/6

y =1

1 + u= 1− u + u2 + O(u3) = 1− u + u2 + O(x3)

u2 = x2/4 + x3/6 + x4/36 = x2/4 + O(x3)x

ex − 1= 1− u + u2 + O(x3)

= 1− x/2− x2/6 + x2/4 + O(x3) = 1− x/2 + x2/12 + O(x3)

More examples


ex = 1 + x + x2/2 + x3/6 + O(x4)

ex − 1 = x + x2/2 + x3/6 + O(x4)

1

y=

ex − 1

x= 1 + x/2 + x2/6 + O(x3) = 1 + u + O(x3) u = x/2 + x2/6

y =1

1 + u= 1− u + u2 + O(u3) = 1− u + u2 + O(x3)

u2 = x2/4 + x3/6 + x4/36 = x2/4 + O(x3)x

ex − 1= 1− u + u2 + O(x3)

= 1− x/2− x2/6 + x2/4 + O(x3)

= 1− x/2 + x2/12 + O(x3)

More examples


ex = 1 + x + x2/2 + x3/6 + O(x4)

ex − 1 = x + x2/2 + x3/6 + O(x4)

1

y=

ex − 1

x= 1 + x/2 + x2/6 + O(x3) = 1 + u + O(x3) u = x/2 + x2/6

y =1

1 + u= 1− u + u2 + O(u3) = 1− u + u2 + O(x3)

u2 = x2/4 + x3/6 + x4/36 = x2/4 + O(x3)x

ex − 1= 1− u + u2 + O(x3)

= 1− x/2− x2/6 + x2/4 + O(x3) = 1− x/2 + x2/12 + O(x3)

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neilstrickland.github.io · Special functions The primary special functions are exp, ln, sin, cos, tan, arcsin, arccos, arctan. Things you should know: I The detailed shape of the