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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERI Outline 1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE POWERS 9 WORKED EXAMPLES

INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

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Page 1: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Outline

1 INTRODUCTION

2 SOLVED EXERCISES

3 THE PRODUCT RULE

4 WORKED EXAMPLES

5 QUOTIENT RULE

6 WORKED EXAMPLES

7 EXERCISES

8 DERIVATIVE OF NEGATIVE POWERS

9 WORKED EXAMPLES

Page 2: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Standard Derivatives

1.

y = xn ;dy

dx= nxn−1

2.

y = ex ;dy

dx= ex

3.

y = ekx ;dy

dx= kekx

4.

y = ln x ;dy

dx=

1

x

5.

y = sin x ;dy

dx= cos x

Page 3: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Standard Derivatives

6.

y = cos x ;dy

dx= − sin x

7.

y = tan x ;dy

dx= sec2 x

8.

y = sec x ;dy

dx= sec x tan x

9.

y = cot x ;dy

dx= −cosec2x

10.

y = cosec2x ;dy

dx= −cosecx cot x

Page 4: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Symbols

dy

dx,

d2y

dx2,

d3y

dx3, · · ·

fx fxx , fxxx , · · ·

f ′(x), f ′′(x), f ′′′(x), · · ·

y ′, y ′′, y ′′′, · · ·

Page 5: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

The General Rule

Giveny = xn

thendy

dx= nxn−1

d

dx(x)

Page 6: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Example

If y = x8, finddy

dx

Page 7: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

dy

dx= 8x7

d

dx(x) = 8x7(1)

= 8x7

Page 8: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

If n is a positive integer, then

d

dx(xn) = nxn−1 · d

dx(x)

Page 9: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Examples

1.d

dx(x5) = 5x4

2.d

dx(x12) = 12x11

3.d

dx(x) = 1x0 = 1

4.d

dx(c) = 0;

d

dx(12) = 0

Page 10: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

5.d

dx(4x8) = 4

d

dx(x8) = 4(8x7) = 32x7

6.

d

dx(x4 − 6x11) =

d

dx(x4)− d

dx(x11)

= 4x3 − 6 · 11x10

= 4x3 − 66x10

Page 11: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Using The Power Rule to find derivvatives of functions

d

dx(X n) = nX n−1 · d

dx(X ), where X is a function.

Given f (u) = Un .Then

d

dx(Un) =

d

dx(Un)

du

dx

= nUn−1 du

dx

Page 12: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Example 1

Findd

dx(x2 − 1)50

Page 13: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

d

dx(x2 − 1)50 = 50(x2 − 1)49

d

dx(x2 − 1)

= 50(x2 − 1)49(2x)

= 100x(x2 − 1)49

Page 14: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Example 2

Findd

dx(x2 − 1)4

Page 15: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

d

dx(x2 − 1)4 = 4(x2 − 1)3

d

dx(x2 − 1)

= 4(x2 − 1)3(2x)

= 8x(x2 − 1)3

Page 16: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Outline

1 INTRODUCTION

2 SOLVED EXERCISES

3 THE PRODUCT RULE

4 WORKED EXAMPLES

5 QUOTIENT RULE

6 WORKED EXAMPLES

7 EXERCISES

8 DERIVATIVE OF NEGATIVE POWERS

9 WORKED EXAMPLES

Page 17: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

EXERCISE 1

y = x13

Solution

dy

dx= 13x12

EXERCISE 2

y = (5x)8

Solution

dy

dx= 8(5x)7(5) = 40(5x)7

Page 18: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Exercise 3

Findy = (x2 + 4)5

Page 19: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

d

dx(x2 + 4)5 = 5(x2 + 4)4

d

dx(x2 + 4)

= 5(x2 + 4)4(2x)

= 10x(x2 + 4)4

Page 20: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Exercise 4

Findy = 3

√(1 + x2)4

Page 21: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

y =(1 + x2

) 43

d

dx=

4

3

(1 + x2

) 13d

dx(1 + x2)

=4

3(1 + x2)

13 (2x)

=8

3(1 + x2)

13

Page 22: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Exercise 5

Findy =

√(2x + 7)

Page 23: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

y = (2x + 7)12

d

dx=

1

2(2x + 7)−

12d

dx(2x)

=1

2(2x + 7)−

12 (2)

= (2x + 7)−12

=1√

2x + 7

Page 24: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Exercise 6

y = (7 + 3x)5,dy

dx=?

Page 25: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

dy

dx= 5(7 + 3x)4

d

dx(7 + 3x)

= 5(7 + 3x)4(3)

= 15(7 + 3x)4

Page 26: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Exercise 7

y = (2x − 3)−2,dy

dx=?

Page 27: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

dy

dx= −2(2x − 3)−3

d

dx(2x − 3)

= −2(2x − 3)4(2)

= −4(2x − 3)−3

= − 4√(2x − 3)3

Page 28: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Exercise 8

y = (3x2 + 5)−3,dy

dx=?

Page 29: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

dy

dx= −3(3x2 + 5)−4

d

dx(3x2 + 5)

= −3(3x2 + 5)−4(6x)

= − 18x√(3x2 + 5)4

Page 30: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Exercise 9

y = (x3 − 3x2 + 7x − 3)4,dy

dx=?

Page 31: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

dy

dx= 4(x3 − 3x2 + 7x − 3)−4

d

dx(x3 − 3x2 + 7x − 3)

= 4(x3 − 3x2 + 7x − 3)3(3x2 − 6x + 7)

Page 32: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Exercise 10

f (x) = 7x5 − 3x4 + 6x2 + 3x + 4, f ′(x) =?

Page 33: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

f ′(x) = 7(5)x4 − 3(4)x3 + 6(2)x + 3

= 35x4 − 12x3 + 12x + 3

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Exercise 10

f (x) = 4x6 + 2x5 − 7x2 + 2x + 5

Page 35: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

fx = 24x5 + 10x5 − 14x + 2

fxx = 120x4 + 40x3 − 13

fxxx = 480x3 + 120x2

Page 36: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Exercises

Finddy

dx

1.y = −8x5 +

√3x3 − 7x

2.y = −7x6 +

√3x2 + 2πx

3.2x50 + 3x12 − 14x2 +

3√

7x +√

5

4.y = x

13

5.(3x2 + 5)−4

Page 37: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

LEVEL 2Exercise 1

If

f (x) = x6 +1

3√x2

, find f ′(x)

Page 38: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

f (x) = x6 + x−23

f ′(x) = 6x5 +

(−2

3

)x−

53

= 6x5 −(

2

3

)x

53

Page 39: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Exercise 2

If

f (x) = 7x3 +10√x

, find f ′(9)

Page 40: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

f (x) = 7x3 + 10x−12

f ′(x) = 21x2 + 10

(−1

2x−

32

)= 21x2 +

(−5x−

32

)= 21x2 − 5x−

32

Page 41: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

f ′(9) = 21(92)− 5(9)−32

= 21(81)− 5

27

= 1708− 5

27= 1700.815

Page 42: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Exercise 3

Find the derivative of y =1

(4x2 − 3)5

Page 43: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

y =1

(4x2 − 3)5= (4x2 − 3)−5

dy

dx= −5(4x2 − 3)−6

d

dx(4x2 − 3)

= −5(4x2 − 3)−6(8x)

= −40x(4x2 − 3)−6

= − 40x

(4x2 − 3)6

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Exercise 4

Find the derivative of y = 3√

3x − 2

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

y = 3√

3x − 2 = (3x − 2)13

dy

dx=

1

3(3x − 2)−

23d

dx(3x − 2)

=1

3(3x − 2)−

23 (3)

= (3x − 2)−23

=1

3√

(3x − 2)2

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Exercise 5

Finddy

dxif y =

3√x− 2√x

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Solution

y =3√x− 2√x = 3x

12 − 2x

12

dy

dx= 3

(−1

2

)x−

32 − 2

(1

2

)x−

12

=

(−3

2

)x−

32 − x−

12

= − 3

2x( 32 )− 1

x( 12 )

= − 3

2√x3− 1√

x

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Exercise 6

Find the derivative of y = (2x5 − 4x3 − x)3

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

y = (2x5 − 4x3 − x)3

dy

dx= 3(2x5 − 4x3 − x)2

d

dy(2x5 − 4x3 − x)

= 3(2x5 − 4x3 − x)2(10x4 − 12x2 − 1)

= 3(10x4 − 12x2 − 1)(2x5 − 4x3 − x)2

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Exercise 7

If y = 1000 + 8x − 150

x2, find

dy

dx

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Solution

y = 1000 + 8x − 150x−2

dy

dx= 0 + 8− 150(−2x−3)

= 8 +300

x3

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Exercise 8

Differentiate f (x) =5

x2− 6

xand find f ′

(1

2

)

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Solution

f (x) = 5x−2 − 6x−1

f ′(x) = −10x−3 + 6x−2

=10

x3+

6

x2

f ′(1

2) = − 10(

1

2

)3 +6(1

2

)2

= −80 + 24

= −56

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Exercise 9

If y =4

3x2 − x + 5,find y ′

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Solution

y ′ = 4d

dx(3x2 − x + 5)−1

= 4(−1)(3x2 − x + 5)−2d

dx(3x2 − x + 5)

= −4(3x2 − x + 5)−2(6x − 1)

=−4(6x − 1)

(3x2 − x + 5)2=

4(1− 6x)

(3x2 − x + 5)2

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Exercise 10

Given y = 3√

(4x2 + 3)2,, finddy

dx

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

dy

dx=

d

dx(4x2 + 3)

23

=2

3(4x2 + 3)−

13 · d

dx(4x2 + 3)

=2

3

(1

4x2 + 3

) 13

(8x)

=16x

3(4x2 + 3)13

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Exercise 11

Find the derivative of y = (6x5 − 4x3 − 5)7

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

y = (6x5 − 4x3 − 5)7

dy

dx= 7(6x5 − 4x3 − 5)6

d

dx(6x5 − 4x3 − 5)

= 7(6x5 − 4x3 − 5)6(30x4 − 12x2)

= 7(30x4 − 12x2)(6x5 − 4x3 − 5)6

= 7(6)(5x4 − 2x2)(6x5 − 4x3 − 5)6

= 42x2(5x2 − 2)(6x5 − 4x3 − 5)6

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Exercise 12

Find the derivative of y =2

4√x3 − x2 − x

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Solution

y =2

4√x3 − x2 − x

= 2(x3 − x2 − x)−14

dy

dx= 2

(−1

4

)(x3 − x2 − x

)− 54d

dy(x3 − x2 − x)

= −1

2

(x3 − x2 − x

)− 54 (3x2 − 2x − 1)

= − (3x2 − 2x − 1)

2(x3 − x2 − x)54

=1 + 2x − 3x2

2( 4√x3 − x2 − x)5

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Exercise 13

d

dx

(x +

1

x

)−3=?

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Solution

d

dx

(x +

1

x

)−3= −3(x +

1

x)−4

d

dx(x +

1

x)

= −3(x +1

x)−4(1− 1

x2)

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Outline

1 INTRODUCTION

2 SOLVED EXERCISES

3 THE PRODUCT RULE

4 WORKED EXAMPLES

5 QUOTIENT RULE

6 WORKED EXAMPLES

7 EXERCISES

8 DERIVATIVE OF NEGATIVE POWERS

9 WORKED EXAMPLES

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

If u and v are differentiable functions of x , thend

dx(uv) = u

dv

dx+ v

du

dx

⇒ If we are asked to differentiate the product of a function:

(a) We may have to keep the first function constant and differentiate thesecond function plus

(b) Keep the second function constant and differentiate the first.

Note: If u, v and w are differentiable functions of x , thend

dy(uvw) = uv

dw

dx+ uw

dv

dx+ vw

du

dx

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Proof:

d

dy(uvw) =

d

dx[(uv)w ] = uv

dw

dx+ w

d

dx(uv)

= uvdw

dx+ w [u

dv

dx+ v

du

dx]

= uvdw

dx+ uw

dv

dx+ vw

du

dx

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Outline

1 INTRODUCTION

2 SOLVED EXERCISES

3 THE PRODUCT RULE

4 WORKED EXAMPLES

5 QUOTIENT RULE

6 WORKED EXAMPLES

7 EXERCISES

8 DERIVATIVE OF NEGATIVE POWERS

9 WORKED EXAMPLES

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Example 1

Differentiate y = (x2 + 1)(x4 − 2x)

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

y = (x2 + 1)(x4 − 2x)

dy

dx= (x2 + 1)

d

dx(x4 − 2x) + (x4 − 2x)

d

dx(x2 + 1)

= (x2 + 1)(4x3 − 2) + (x4 − 2x)(2x)

= 4x5 − 2x2 + 4x3 − 2 + 2x5 − 4x2

= 6x5 + 4x3 − 6x2 − 2

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Example 2

d

dx

(x +

1

x

)−3=?

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

d

dx

(x +

1

x

)−3= −3(

(x +

1

x

)−4d

dx

(x +

1

x

))

= −3(

(x +

1

x

)−4(1− 1

x2

)

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Example 3

Finddy

dxif y = x5(3x3 − 2x + 1)3

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Solution

y = x5(3x3 − 2x + 1)3

= x5d

dx(3x3 − 2x + 1)3 + (3x3 − 2x + 1)3

d

dx(x5)

= x5 · 3(3x3 − 2x + 1)2d

dx(3x3 − 2x + 1) + (3x3 − 2x + 1)3(5x4)

= 3x5(3x3 − 2x + 1)2(9x2 − 2) + 5x4(3x3 − 2x + 1)3

= x4(3x3 − 2x + 1)2[3x(9x2 − 2) + 5(3x3 − 2x + 1)

]= x4(3x3 − 2x + 1)2(27x3 − 6x + 15x3 − 10x + 5)

= x4(3x3 − 2x + 1)2(42x3 − 16x + 5)

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Example 4

Calculatedy

dxif y = (2x4 − 1)4(x8 − 3x2)7

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

y = (2x4 − 1)4(x8 − 3x2)7

dy

dx= (2x4 − 1)4

d

dx(x8 − 3x2)7 + (x8 − 3x2)7

d

dx(2x4 − 1)4

= (2x4 − 1)4 · 7(x8 − 3x2)6dy

dx(x8 − 3x2)

+ (x8 − 3x2)7 · 4(2x4 − 1)3dy

dx(2x4 − 1)

= 7(2x4 − 1)4(x8 − 3x2)6(8x7 − 6x) + 4(2x4 − 1)3(x8 − 3x2)7(8x3)

= (2x4 − 1)3(x8 − 3x2)6[7(2x4)(8x7 − 6x) + 32x3(x8 − 3x2)

]= (2x4 − 1)3(x8 − 3x2)6(112x11 − 84x5 − 56x7 + 42x + 23x11− 96x5)

= (2x4 − 1)3(x8 − 3x2)6(144x11 − 56x7 − 180x5 + 42x)

= (2x4 − 1)3(x8 − 3x2)6 · x(144x10 − 56x6 − 180x4 + 42)

= x(2x4 − 1)3(x8 − 3x2)6(144x10 − 56x6 − 180x4 + 42)

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Example 5

Finddy

dxif y = (2x3 − 1)5

√x − 1

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

dy

dx= (2x3 − 1)5

d

dx(x − 1)

12 + (X − 1)

12d

dx(2x3 − 1)5

= (2x3 − 1)51

2((x − 1)−

12 )

d

dx(x − 1) + (x − 1)

12 · 5(2x3 − 1)4

d

dx(2x3 − 1)

= (2x3 − 1)5 · 1

2((x − 1)−

12 )(1) + (x − 1)

12 · 5(2x3 − 1)4(6x2)

=1

2(2x3 − 1)5 · ((x − 1)−

12 ) + (x − 1)

12 (2x3 − 1)4(30x2)

=

1

2(2x3 − 1)5

(x − 1)12

+ (30x2)(x − 1)12 (2x3 − 1)4

=(2x3 − 1)5

2√x − 1

+ (30x2)(2x3 − 1)4√x − 1

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Example 6

Finddy

dxif y = x2(2x − 1)(6x + 5)

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

dy

dx= x2(2x − 1)

d

dx(6x + 5) + x2(6x + 5)

d

dx(2x − 1)

+ (2x − 1)(6x + 5)d

dx(x2)

= x2(2x − 1)(6) + x2(6x + 5)(2) + (2x − 1)(6x + 5)(2x)

= 6x2(2x − 1) + 2x2(6x + 5) + (2x − 1)(12x2 + 10x)

= 48x3 + 12x2 − 10x

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Example 7

Given that y = (3x − 1)2(2x2 − 3)(x3 + 4)5 ;dy

dx=?

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

dy

dx= (3x − 1)2(2x2 − 3)

d

dx(x3 + 4)5 + (3x − 1)2(x3 + 4)5

d

dx(2x2 − 3)

+ (2x2 − 3)(x3 + 4)5d

dx(3x − 1)2

= (3x − 1)2(2x2 − 3) · 5(x3 + 4)4d

dx(x3 + 4) + (3x − 1)2(x3 + 4)5(4x)

+ (2x2 − 3)(x3 + 4)5(2)(3x − 1)d

dx(3x − 1)

= (3x − 1)2(2x2 − 3) · 5(x3 + 4)4(3x2) + 4x(3x − 1)2(x3 + 4)5

+ 2(2x2 − 3)(x3 + 4)5(3x − 1) · 3= 15x2(3x − 1)2(2x2 − 3)(x3 + 4)4 + 4x(3x − 1)2(x3 + 4)5

+ 6(2x2 − 3)(x3 + 4)5(3x − 1)

= (3x − 1)(x3 + 4)4[15x2(2x2 − 3)(2x + 1) + 4x(3x − 1)(x3 + 4)

+ 6(2x2 − 3)(x3 + 4)]

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Example 8

Finddy

dxif y = (8x3 − x2)5(7x2 − 3x)2(x5 − 2)10

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

dy

dx= (8x3 − x2)5(7x2 − 3x)2

d

dx(x5 − 2)10

+ (8x3 − x2)5(x5 − 2)10d

dx(7x2 − 3x)2

+ (7x2 − 3x)2(x5 − 2)10d

dx(8x3 − x2)5

dy

dx= (8x3 − x2)5(7x2 − 3x)2 · 10(x5 − 2)9

d

dx(x5 − 2)

+ (8x3 − x2)5(x5 − 2)10 · 2(7x2 − 3x)d

dx(7x2 − 3x)

+ (7x2 − 3x)2(x5 − 2)10 · 5(8x3 − x2)4d

dx(8x3 − x2)

dy

dx= (8x3 − x2)5(7x2 − 3x)2 · 10(x5 − 2)9 · 5x4

+ (8x3 − x2)5(x5 − 2)10 · 2(7x2 − 3x)(14x − 3)

+ (7x2 − 3x)2(x5 − 2)10 · 5(8x3 − x2)4(24x2 − 2x)

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

dy

dx= 50x4(7x2 − 3x)2(8x3 − x2)5(x5 − 2)9

+ 2(8x3 − x2)5(x5 − 2)10(7x2 − 3x)(14x − 3)

+ 5(7x2 − 3x)2(x5 − 2)10(8x3 − x2)4(24x2 − 2x)

dy

dx= (7x2 − 3x)(8x3 − x2)4(x5 − 2)9[50x4(8x3 − x2)(7x2 − 3x)

+ 2(x5 − 2)(8x3 − x2)(14x − 3) + 5(x5 − 2)(7x2 − 3x)2(24x2 − 2x)]

dy

dx= (7x2 − 3x)(8x3 − x2)4(x5 − 2)9[(400x7 − 50x6)(7x2 − 3x)

+ (2x5 − 4)(8x3 − x2)(14x − 3) + (5x5 − 10)(7x2 − 3x)2(24x2 − 2x)]

Page 85: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Example 9

Finddy

dxif y = (2x5 − 3x3)4(5x − 4x2)3(x6 − 6x2)7

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

dy

dx= (2x5 − 3x3)4(5x − 4x2)3

d

dx(x6 − 6x2)7

+ (2x5 − 3x3)4(x6 − 6x2)7d

dx(5x − 4x2)3

+ (5x − 4x2)3(x6 − 6x2)7d

dx(2x5 − 3x3)4

dy

dx= (2x5 − 3x3)4(5x − 4x2)3 · 7(x6 − 6x2)6

d

dx(x6 − 6x2)

+ (2x5 − 3x3)4(x6 − 6x2)7 · 3(5x − 4x2)2d

dx(5x − 4x2)

+ (5x − 4x2)3(x6 − 6x2)7 · 4(2x5 − 3x3)3d

dx(2x5 − 3x3)

dy

dx= 7(2x5 − 3x3)4(5x − 4x2)3(x6 − 6x2)6(6x5 − 12x)

+ 3(2x5 − 3x3)4(x6 − 6x2)7(5x − 4x2)2(5− 8x)

+ 4(5x − 4x2)3(x6 − 6x2)7(2x5 − 3x3)3(10x4 − 9x2)

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

dy

dx= 7(6x5 − 12x)(2x5 − 3x3)4(5x − 4x2)3(x6 − 6x2)6

+ 3(5− 8x)(2x5 − 3x3)4(x6 − 6x2)7(5x − 4x2)2

+ 4(2x5 − 3x3)3(10x4 − 9x2)(5x − 4x2)3(x6 − 6x2)7

dy

dx= (2x5 − 3x3)3(5x − 4x2)2(x6 − 6x2)6[7(6x5 − 12x)(2x5 − 3x3)(5x − 4x2)

+ 3(5− 8x)(2x5 − 3x3)4(x6 − 6x2) + 4(10x4 − 9x2)(5x − 4x2)(x6 − 6x2)]

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Note:If f and g are differentiable at x , thend

dx[f (x)g(x)] = f (x)

d

dx[g(x)] + g(x)

d

dx[f (x)]

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Example 10

Finddy

dxif y = (4x2 − 1)(7x3 + x)

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

dy

dx=

d

dx[(4x2 − 1)(7x3 + x)]

= (4x2 − 1)d

dx(7x3 + x) + (7x3 + x)

d

dx(4x2 − 1)

= (4x2 − 1)(21x2 + 1) + (7x3 + x)(8x)

= 140x4 − 9x2 − 1

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Example 11

If f (x) = (x2 + x)(2x3 − 3) find f ′(x)

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

f ′(x) = (x2 + x)d

dx(2x3 − 3) + (2x3 − 3)

d

dx(x2 + x)

= (x2 + x)(6x2) + (2x3 − 3)(2x + 1)

= 10x4 + 8x2 − 6x − 3

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Example 12

If g(x) = x15 (x − 1)

35 , find g ′(x)

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

g ′(x) = x15 · 3

5(x − 1)−

25 + (x − 1)

35 · 1

5x−

45

=3x

15

5(x − 1)25

+(x − 1)

35

5x45

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Example 13

f (x) = x2(1− 3x3)13 , f ′(x) =?

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

f ′(x) = x2d

dx(1− 3x3)

13 + (1− 3x3)

13

(d

dx(x2)

)= x2

[1

3(1− 3x3)−

23 )

d

dx(1− 3x3)

]+ (1− 3x3)

13 (2x)

=1

3x2(1− 3x3)−

23 )(−9x2) + 2x(1− 3x3)

13

= −3x4(1− 3x3)−23 + 2x(1− 3x3)

13

Page 97: INTRODUCTIONSOLVED EXERCISESTHE PRODUCT ......1 INTRODUCTION 2 SOLVED EXERCISES 3 THE PRODUCT RULE 4 WORKED EXAMPLES 5 QUOTIENT RULE 6 WORKED EXAMPLES 7 EXERCISES 8 DERIVATIVE OF NEGATIVE

INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

= x(1− 3x3)−23

[−3x2 + 2(1− 3x3)

]= x(1− 3x3)−

23 (2− 9x3)

=x(2− 9x3)

(1− 3x3)−23

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Outline

1 INTRODUCTION

2 SOLVED EXERCISES

3 THE PRODUCT RULE

4 WORKED EXAMPLES

5 QUOTIENT RULE

6 WORKED EXAMPLES

7 EXERCISES

8 DERIVATIVE OF NEGATIVE POWERS

9 WORKED EXAMPLES

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

If f (x) =u(x)

v(x), where u and v are differentiable functions of x , then

f ′(x) =v(x)u′(x)− u(x)v ′(x)

[v(x)]2

where v(x) 6= 0

Thus if u and v are differentiable function of x and y =u

v, where v 6= 0,

then

dy

dx= y ′ =

vdu

dx− u

dv

dxv2

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Outline

1 INTRODUCTION

2 SOLVED EXERCISES

3 THE PRODUCT RULE

4 WORKED EXAMPLES

5 QUOTIENT RULE

6 WORKED EXAMPLES

7 EXERCISES

8 DERIVATIVE OF NEGATIVE POWERS

9 WORKED EXAMPLES

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Example 1

If y =x + 1

x − 1, find

dy

dx

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

dy

dx=

(x − 1)d

dx(x + 1)− (x + 1)

d

dx(x − 1)

(x − 1)2

=(x − 1)(1)− (x + 1)(1)

(x − 1)2=

x − 1− x − 1

(x − 1)2

=−2

(x − 1)2= − 2

(x − 1)2

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Example 2

Finddy

dxif y =

x3

x2 + 1

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

dy

dx=

(x2 + 1)d

dx(x3)− x3

d

dx(x2 + 1)

(x2 + 1)2

=(x2 + 1)(3x2)− x3(2x)

(x2 + 1)2

=3x4 + 3x2 − 2x4

(x2 + 1)2

=x4 + 3x2

(x2 + 1)2

=x2(x2 + 3)

(x2 + 1)2

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Example 3

Calculate y ′ when y =2x5 − x3

3x4 − 2x

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

dy

dx=

(3x4 − 2x)d

dx(2x5 − x3)− (2x5 − x3)

d

dx(3x4 − 2x)

(3x4 − 2x)2

=(3x4 − 2x)(10x4 − 3x2)− (2x5 − x3)(12x3 − 2)

(3x4 − 2x)2

=30x8 − 9x6 − 20x5 + 6x3 − 24x8 + 4x5 + 12x6 − 2x3

(3x4 − 2x)2

=6x8 + 3x6 − 16x5 + 4x3

(3x4 − 2x)2

=x3(6x5 + 3x3 − 16x2 + 4)

(3x4 − 2x)2

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Example 4

Finddy

dxif y =

(3x2 − x)4

(x3 − 2x4)3

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

dy

dx=

(x3 − 2x4)3d

dx(3x2 − x)4 − (3x2 − x)4

d

dx(x3 − 2x4)3

(x3 − 2x4)3·2

=(x3 − 2x4)34(3x2 − x)3

d

dx(3x2 − x)− (3x2 − x)43(x3 − 2x4)

d

dx(x3 − 2x4)

(x3 − 2x4)6

=4(x3 − 2x4)3(3x2 − x)3(6x − 1)− 3(3x2 − x)4(x3 − 2x4)(3x2 − 8x3)

(x3 − 2x4)6

=(x3 − 2x4)2(3x2 − x)3

[4(6x − 1)(x3 − 2x4)− 3(3x2 − 8x3)(3x2 − x)4

](x3 − 2x4)6

=(3x2 − x)3

(x3 − 2x4)4(24x5 − 19x4 + 5x3)

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Example 5

Finddy

dxif y =

(2x − 3

5x2 + 1

)7

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

dy

dx= 7

(2x − 3

5x2 + 1

)6

(5x2 + 1)d

dx(2x − 3)− (2x − 3)

d

dx(5x2 + 1)

(5x2 + 1)2

= 7

(2x − 3

5x2 + 1

)6 [(5x2 + 1)(2)− (2x − 3)(10x)

(5x2 + 1)2

]= 7

(2x − 3

5x2 + 1

)6 [10x2 + 2− 20x2 + 30x

(5x2 + 1)2

]= 7

(2x − 3

5x2 + 1

)6(2 + 30x − 10x2

(5x2 + 1)2

)=

7(2x − 3)6(2 + 30x − 10x2)

(5x2 + 1)6(5x2 + 1)2

=14(1 + 15x − 5x2)(2x − 3)6

(5x2 + 1)8

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Example 6

Calculated

dx

[x3 − 1

5√

4x3 + 3

]

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

d

dx

[x3 − 1

5√

4x3 + 3

]=

5√

4x3 + 3d

dx(x3 − 1)− d

dx(4x2 + 3)

15

( 5√

4x3 + 3)2

=

5√

4x3 + 3(3x2)− (x3 − 1) · 1

5(4x2 + 3)−

45d

dx(4x2 + 3)

( 5√

4x3 + 3)2

=3x2 5√

4x3 + 3− 1

5(x3 − 1)(4x2 + 3)−

45 (8x)

( 5√

4x3 + 3)2

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Solution

=3x2 5√

4x3 + 35√

4x3 + 3)2− 8x(x3 − 1)

5· 1

5√

4x3 + 3)2· 1

(4x2 + 3)

4

5

=3x2

5√

(4x3 + 3)− 8x(x3 − 1)

5· 1

5√

4x3 + 3)2· 1

5√

(4x3 + 3)4

=3x2

5√

(4x3 + 3)− 8x(x3 − 1)

5 5√

(4x3 + 3)6

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Example 7

Find y ′ when y =

√x − 1

x3

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Solution

dy

dx=

x3d

dx(x − 1)

12 − (x − 1)

12d

dx(x3)

(x3)2

=x3 · 1

2(x − 1)−

12d

dx(x − 1)− (x − 1)

12 (3x2)

x6

=

1

2x3(x − 1)−

12 (1)− (x − 1)

12 (3x2)

x6

=

1

2x3(x − 1)−

12 − (x − 1)

12 (3x2)

x6

=x3 − 6x2(x − 1)

2x6(x − 1)12

=6x2 − 5x3

2x6√x − 1

=x2(6− 5x)

2x6√x − 1

=(6− 5x)

2x4√x − 1

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Example 8

If g(t) =t3

1− t4find g ′(t)

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Solution

g ′(t) =(1− t4)(3t2)− t3(−4t3)

(1− t4)2

=3t2 − 3t6 + 4t6

(1− t4)2

=3t2 + t6

(1− t4)2

=t2(3 + t4)

(1− t4)2

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Example 9

Given that f (x) =2x + 3

x2 + 1, find f ′(x)

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Solution

f ′(x) =(x2 + 1)(2)− (2x + 3)(2x)

(x2 + 1)2

=2x2 + 2− 4x2 − 6x

(x2 + 1)2

=2x2 − 6x + 2

(x2 + 1)2

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EXERCISES

1.f (x) = x2(3x2 − 1)

2.y = 6x4(x3 + 2x)

3.f (x) = (1− x2)(1 + x2)

4.

g(x) =1

x3(x2 − x)

5.

y =

(2

x+ 1

)(1

x2− 3

)

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EXERCISES

6.g(x) = (3x4 − 1)(5x2 − 6x + 11)

7.s = (t2 + 1)(2t2 − t + 3)

8.y = (1−

√x)(2 +

√x)

9.

f (x) =2

x3

10.y =

x

x + 1

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Outline

1 INTRODUCTION

2 SOLVED EXERCISES

3 THE PRODUCT RULE

4 WORKED EXAMPLES

5 QUOTIENT RULE

6 WORKED EXAMPLES

7 EXERCISES

8 DERIVATIVE OF NEGATIVE POWERS

9 WORKED EXAMPLES

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EXERCISES

11.

y =2x

x + 1

12.

f (x) =x − 3

2x + 5

13.

g(x) =5x + 1

x2 − 1

14.

y =x2

x3 + 4

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15.

y =x2 − x + 1

x2 + 2x − 3

16.

g(x) =7x4 + 11

x − 2

17.

f (x) =

(1 +

1

x

)(1 +

1

x2

)

18.

f (x) =(9x8 − 8x9

)(x +

1

x

)

19.

g(x) =2x2 + 1

x + 5

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20.

f (x) =x4

4− x3

3+

x2

2

21.

y =4x3 + 1

x3 − 1

22.

y =

(x

1 + x

)(2− x

3

)

23.

f (x) = (2x − 3)

(2x − 3

x

)

24.

f (x) = (7x3 − 4x2 + 2)

1

4

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INTRODUCTION SOLVED EXERCISES THE PRODUCT RULE WORKED EXAMPLES QUOTIENT RULE WORKED EXAMPLES EXERCISES DERIVATIVE OF NEGATIVE POWERS WORKED EXAMPLES

Outline

1 INTRODUCTION

2 SOLVED EXERCISES

3 THE PRODUCT RULE

4 WORKED EXAMPLES

5 QUOTIENT RULE

6 WORKED EXAMPLES

7 EXERCISES

8 DERIVATIVE OF NEGATIVE POWERS

9 WORKED EXAMPLES

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For each negative integer n ,f (x) = xn has the derivativef ′(x) = nxn−1

In particular

f (x) = x−1; f ′(x) = (−1)x−2 = −x−2

f (x) = x−2; f ′(x) = (−2)x−3 = − 2

x3

f (x) = x−3; f ′(x) = (−3)x−3 = − 3

x4

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Example

If f (x) =5

x2− 6

x, find f ′(x) and f ′(

1

2)

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Solution

f (x) = 5x−2 − 6x−1

f ′(x) = 5(2)x−3 − 6(−1)x−2

= −10

x3+

6

x2

f ′(1

2) = − 10

(1

2)3

+6

(1

2)2

= −80 + 24 = −56

Note: Because y =1

xncan be expressed as y = x−n, all problems where

quotient rule is applicable, can be converted to products so that youapply product rule. Here, it is a matter of choice or preference.

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Outline

1 INTRODUCTION

2 SOLVED EXERCISES

3 THE PRODUCT RULE

4 WORKED EXAMPLES

5 QUOTIENT RULE

6 WORKED EXAMPLES

7 EXERCISES

8 DERIVATIVE OF NEGATIVE POWERS

9 WORKED EXAMPLES

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Example 1

Differentiate :(i)(x2 − 2)(x + 3)−2 as a product

(ii)x2 − 2

(x + 3)2as a quotient

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Solution (i)

Let y = (x2 − 2)(x + 3)−2

dy

dx= (x2 − 2)

d

dx(x + 3)−2 + (x + 3)−2

d

dx(x2 − 2)

= (x2 − 2) · −2(x + 3)−3d

dx(x + 3) + (x + 3)−2(2x)

=−2(x2 − 2)

(x + 3)3(1) +

2x

(x + 3)2

=−2x2 + 4 + 2x(x + 3)

(x + 3)3

=−2x2 + 4 + 2x2 + 6x

(x + 3)3

=4 + 6x

(x + 3)3=

2(2 + 3x)

(x + 3)3

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Solution (ii)

y =x2 − 2

(x + 3)2

dy

dx=

(x + 3)2d

dx(x2 − 2)− (x2 − 2)

d

dx(x + 3)2

[(x + 3)2]2

=2x(x + 3)2 − 2(x + 3)(x2 − 2)(1)

(x + 3)4

=(x + 3)

[2x(x + 3)− 2(x2 − 2)

](x + 3)4

=2x2 + 6x − 2x2 + 4

(x + 3)3=

6x + 4

(x + 3)3=

2(2 + 3x)

(x + 3)3

NB: On comparing the results of (i) and (ii), you will notice that final

results are the same . Hence , y =x2 − 2

(x + 3)2= (x2 − 2)(x + 3)2

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Example 2

Differentiate :(i)(x − 1)3(x3 − 1)−1 as a product

(ii)(x − 1)3

(x3 − 1)as a quotient

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Solution (i)

y = (x − 1)3(x3 − 1)−1

dy

dx= (x − 1)3

d

dx(x3 − 1)−1 + (x3 − 1)−1

d

dx(x − 1)3

= (x − 1)3 · (−1)(x3 − 1)−2d

dx(x3 − 1) + (x3 − 1)−1 · 3(x − 1)2

d

dx(x − 1)

= −(x − 1)3(x3 − 1)−2(3x2) + 3(x3 − 1)−1(x − 1)2(1)

=−3x2(x − 1)3

(x3 − 1)2+

3(x − 1)2

(x3 − 1)

=−3x2(x − 1)3 + (x3 − 1)2

(x3 − 1)2

=(x − 1)2

[−3x2(x − 1) + 3(x3 − 1)

](x3 − 1)2

=(x − 1)2(−3x2 + 3x2 + 3x3 − 3)

(x3 − 1)2=

(3x2 − 3(x − 1)2)

(x3 − 1)2=

3(x + 1)(x − 1)3

(x3 − 1)2

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Solution (ii)

y =(x − 1)3

(x3 − 1)

dy

dx=

(x3 − 1)d

dx(x − 1)3 − (x − 1)3

d

dx(x3 − 1)

(x3 − 1)2

=(x3 − 1) · 3(x − 1)2

d

dx(x − 1)− (x − 1)3(3x2)

(x3 − 1)2

=3(x3 − 1)(x − 1)2(1)− (x − 1)3(3x2)

(x3 − 1)2

=3(x − 1)2[(x3 − 1)− x2(x − 1)]

(x3 − 1)2

=3(x − 1)2[(x3 − 1)− x2(x − 1)]

(x3 − 1)2

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EXERCISE

Differentiate :(i) as a product(ii) as a quotient