MS121: IT Mathematics€¦ · mPQ = f(x) f(a) x a Di erentiation (1/5) MS121: IT Mathematics John Carroll 8 / 58. Secant Lines & Tangents Tangents Di erentiation: An Introduction

MS121: IT Mathematics

Differentiation

Introduction

John CarrollSchool of Mathematical Sciences

Dublin City University

Outline

1 Introduction

2 Secant Lines & Tangents

3 Differentiation from First Principles

4 First Principles: Examples

5 How Can a Function Fail to Be Differentiable?

Introduction

Outline

1 Introduction





Differentiation (1/5) MS121: IT Mathematics John Carroll 3 / 58

Introduction Differentiation

Differentiation

What is Differentiation

Differentiation is a mathematical technique for analysing the way inwhich functions change.

In particular, it determines how rapidly a function is changing at anyspecific point.

Differentiation also allows us to find maximum and minimum valuesof a function which can be useful in determining optimum values ofkey variables.



Differentiation from First Principles

Slope of a Line

Recall how we find the slope of a straight line connecting two points(x1, y1) and (x2, y2), given by

slope =y2 − y1

x2 − x1.

We could also write this as

slope =change in y

change in x=

dy

dx




Slope of a Curve?

How do we define the slope of a curve? We define it in terms of the slopeof a straight line which we already know.

Finding the Slope of a Curve

Take any curve and the point at which the slope is required.

At this point, draw the tangent to the curve.

The tangent is a straight line which, at the point in question, appearsto be parallel to the curve.

The slope of the curve at this point is simply the slope of the tangent.


Secant Lines & Tangents

Outline

1 Introduction






Secant Lines & Tangents Tangents

Differentiation: An Introduction — Tangents

If a curve C has equation y = f (x)and we want to find the tangentline to C at the point P(a, f (a)),then we consider a nearby pointQ(x , f (x)), where x 6= a, andcompute the slope of the secantline:

mPQ =f (x)− f (a)

x − a




mPQ =f (x)− f (a)

x − a

Then we let Q approach Palong the curve C by letting xapproach a.

If mPQ approaches a number m,then we define the tangent t tobe the line through P with slopem.

The tangent line is the limitingposition of the secant line PQas Q approaches P.




Definition

The tangent line to the curvey = f (x) at the point at the pointP(a, f (a)) is the line through P withslope

m = limx→a

f (x)− f (a)

x − a

provided that this limit exists.




Easier Expression

If h = x − a, then x = a + h andso the slope of the secant linePQ is

mPQ =f (a + h)− f (a)

h

As x approaches a, h approaches0 (because h = x − a) and sothe expression for the slope ofthe tangent line becomes

m = limh→0

f (a + h)− f (a)

h


Secant Lines & Tangents Illustration using Secant Lines

Tangent to y = x2 at the point x = 1

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−1

−0.5

0

0.5

1

1.5

2

2.5

3

3.5

4

x

y

y = x2

y = 2x − 1




−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−1

−0.5

0

0.5

1

1.5

2

2.5

3

3.5

4

x

y

y = x2

y = 2x − 1

Slope of a Curve

As you will see from the graph, the slopes at different points on the curvewill be different (unlike the straight line where the slope is always thesame, no matter where it is evaluated).



Limit of Secant Lines: y = x2 at the point x = 1

SECANT LINE ILLUSTRATION

f(x) = x2 C = (1,1)

−1.5 −1 −0.5 0 0.5 1 1.5 2 2.5−0.5

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

*C

o

o

Secant slopes.

1.4695





f(x) = x2 C = (1,1)

−1.5 −1 −0.5 0 0.5 1 1.5 2 2.5−0.5

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

*C

o

o

Secant slopes.

1.4695

1.5453





f(x) = x2 C = (1,1)

−1.5 −1 −0.5 0 0.5 1 1.5 2 2.5−0.5

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

*C

o

o

Secant slopes.

1.4695

1.5453

1.6211





f(x) = x2 C = (1,1)

−1.5 −1 −0.5 0 0.5 1 1.5 2 2.5−0.5

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

*C

o

o

Secant slopes.

1.4695

1.5453

1.6211

1.6969





f(x) = x2 C = (1,1)

−1.5 −1 −0.5 0 0.5 1 1.5 2 2.5−0.5

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

*C

o

o

Secant slopes.

1.4695

1.5453

1.6211

1.6969

1.7726





f(x) = x2 C = (1,1)

−1.5 −1 −0.5 0 0.5 1 1.5 2 2.5−0.5

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

*C

o

o

Secant slopes.

1.4695

1.5453

1.6211

1.6969

1.7726

1.8484





f(x) = x2 C = (1,1)

−1.5 −1 −0.5 0 0.5 1 1.5 2 2.5−0.5

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

*C

o

o

Secant slopes.

1.4695

1.5453

1.6211

1.6969

1.7726

1.8484

2

Slope of Tangent Line



Example

Find an equation of the tangent lineto the hyperbola y = 3

x at the point(3, 1).



Solution (1/2): Slope of Tangent

Let f (x) = 3x . Then the slope of the

tangent at (3, 1) is:

m = limh→0

f (3 + h)− f (3)

h

= limh→0

33+h − 1

h

= limh→0

−1

3 + h

= −1

3



Solution (2/2): Equation of Tangent

Therefore, an equation of thetangent at the point (3, 1) is

y − 1 = −1

3(x − 3)

which simplifies to

x + 3y − 6 = 0



Outline

1 Introduction






Differentiation from First Principles Definition


Method of Differentiation

We do not need to keep drawing tangents to find the slopes — wecan use differentiation.

The derivative (obtained by differentiation) of a curve at a point isthe slope of the curve at that point.

We shall first differentiate functions from first principles.



The Derivative of a Function

Definition

The derivative of the function f : R → R at a value x = a, if it exists, is

limh→0

f (a + h)− f (a)

h




Basic Approach

The idea is to consider the line joining the two points (a, f (a)) and(a + h, f (a + h)) which, for h small, will approximate the tangent tothe curve at the point x = a.

Compute the slope of this line:

slope =y2 − y1

x2 − x1=

f (a + h)− f (a)

(a + h)− a=

f (a + h)− f (a)

h

and then take the limit as h tends to zero, giving

dy

dx= lim

h→0

f (a + h)− f (a)

h.





First Principles: Examples

Outline

1 Introduction






First Principles: Examples Using the Limit Equation

Example 1

We will find the derivative from first principles of f (x) = x2 at x = a.

Solution

We will find the derivative from first principles of f (x) = x2 at x = a.First compute

f (a + h)− f (a)

h=

(a + h)2 − a2

h

=(a2 + 2ah + h2)− a2

h

=2ah + h2

h= 2a + h



f (x) = x2

Example 1 (Cont’d)

Then take limits to obtain

limh→0

f (x + h)− f (x)

h= lim

h→0(2a + h) = 2a



The function y = x2 on the interval [0, 3]

0 0.5 1 1.5 2 2.5 30

1

2

3

4

5

6

7

8

9

y = x2



Function y = x2 1st Secant Line with h = 2

0 0.5 1 1.5 2 2.5 30

1

2

3

4

5

6

7

8

9

y = x2



Function y = x2 2nd Secant Line with h = 1.75

0 0.5 1 1.5 2 2.5 30

1

2

3

4

5

6

7

8

9

y = x2



Function y = x2 3rd Secant Line with h = 1.5

0 0.5 1 1.5 2 2.5 30

1

2

3

4

5

6

7

8

9

y = x2



Function y = x2 4th Secant Line with h = 1.25

0 0.5 1 1.5 2 2.5 30

1

2

3

4

5

6

7

8

9

y = x2



Function y = x2 5th Secant Line with h = 1

0 0.5 1 1.5 2 2.5 30

1

2

3

4

5

6

7

8

9

y = x2



Function y = x2 6th Secant Line with h = 0.75

0 0.5 1 1.5 2 2.5 30

1

2

3

4

5

6

7

8

9

y = x2



Function y = x2 Limit as h→ 0

0 0.5 1 1.5 2 2.5 30

1

2

3

4

5

6

7

8

9

x

y

y = x2

y = 2x − 1



Differentiation

Notation

The notation we use is as follows:

limh→0

f (a + h)− f (a)

h=

df (x)

dx

∣∣∣∣x=a

=dy

dx

∣∣∣∣x=a

where the vertical line (above) denotes “evaluated at”.



Differentiation

Other Notation

It is important to note at this point that an alternative (“∆x”) notation isalso widely used, namely:

lim∆x→0

(y + ∆y)− y

(x + ∆x)− x= lim

∆x→0

∆y

∆x

=dy

dx.

We illustrate the use of the “∆” in the following example.



Example 2

Differentiate the function y = x3 + 1 from first principles at x = a.

Solution

First compute the quotient ∆y∆x :

y = x3 + 1

y + ∆y = (x + ∆x)3 + 1

∆y = (x + ∆x)3 − x3

= x3 + 3x2∆x + 3x(∆x)2 + (∆x)3 − x3

= 3x2∆x + 3x(∆x)2 + (∆x)3

∆y

∆x= 3x2 + 3x∆x + (∆x)2



y = x3 + 1


We then take the limit as ∆x tends to zero:

∆y

∆x= 3x2 + 3x∆x + (∆x)2

dy

dx= lim

∆x→0

∆y

∆x

= 3x2

⇒ dy

dx

∣∣∣∣x=a

= 3a2



Example 3

As a simpler example, we will find the derivative from first principles off (x) = x .

Solution

We compute

f (a + h)− f (a)

h=

(a + h)− a

h=

h

h= 1

so that, irrespective of h, we obtain

limh→0

f (x + h)− f (x)

h= lim

h→01 = 1

This is as expected since y = x is just the straight line with constant slope(= 1).



Example 4

Differentiate from first principles the function y =√x for x > 0.

Solution

Since

f (x + h)− f (x)

h=

√x + h −

√x

h

=

√x + h −

√x

h·[√

x + h +√x√

x + h +√x

]=

1√x + h +

√x

we obtain

limh→0

f (x + h)− f (x)

h= lim

h→0

1√x + h +

√x

=1

2√x



y =√x x > 0

Solution (Cont’d)

We then obtain

limh→0

f (x + h)− f (x)

h= lim

h→0

1√x + h +

√x

=1

2√x



y =√x x > 0



Footnote to Example 4

Note that, in order to simplify the expression√x + h −

√x , we

multiplied above and below by the “conjugate” expression√x + h +

√x .

We are making use of the identity

A2 − B2 = (A− B)(A + B)

and on the basis that it may sometimes be easier to deal withA2 − B2 than A− B (as in this case).



Example 5

What is the slope of the curve y = 1x at the point x = 2?

Solution

As before, we compute:

f (x + h)− f (x)

h=

1x+h −

1x

h

=x − (x + h)

x(x + h)h

=−h

x(x + h)h

=−1

x(x + h)



y =1

x


Then take limits to obtain

limh→0

f (x + h)− f (x)

h=

−1

x(x + 0)= − 1

x2.

Finally, evaluate at x = 2 to obtain

dy

dx

∣∣∣∣x=2

= − 1

x2

∣∣∣∣x=2

= −1

4.


How Can a Function Fail to Be Differentiable?

Outline

1 Introduction






How Can a Function Fail to Be Differentiable? Definition

When is a Function Differentiable?

Definition

A function is differentiable at a if f ′(a) exists.

It is differentiable on an open interval (a, b) if it is differentiable atevery number in the interval.


How Can a Function Fail to Be Differentiable? Example: |x|


Example: f (x) = |x |




Example: f (x) = |x |: Case x > 0

If x > 0, then |x | = x and we can choose h small enough so thatx + h > 0 and hence |x + h| = x + h. Therefore, for x > 0, we have

f ′(x) = limh→0

|x + h| − |x |h

= limh→0

x + h − x

h

= limh→0

h

h= lim

h→01 = 1

and so f is differentiable for any x > 0.




Example: f (x) = |x |: Case x < 0

If x < 0, then |x | = −x and we can choose h small enough so thatx + h < 0 and hence |x + h| = −(x + h). Therefore, for x < 0, we have

f ′(x) = limh→0

−(x + h)− (−x)

h

= limh→0

−x − h + x

h

= limh→0

−hh

= limh→0−1 = −1

and so f is differentiable for any x < 0.




f ′(0) does not exist.




A formula for f ′ is given by

f ′(x) =

{−1 if x < 0+1 if x > 0

Note

f (x) = |x | is differentiable at all x except 0.


How Can a Function Fail to Be Differentiable? Summary of Cases

How Can a Function Fail to Be Differentiable?

If the graph of a function f has a “corner” or “kink” in it, then the graphof has no tangent at this point and is not differentiable there.

The Three Cases


Documents

MS121: IT Mathematics€¦ · mPQ = f(x) f(a) x a Di erentiation (1/5) MS121: IT Mathematics John Carroll 8 / 58. Secant Lines & Tangents Tangents Di erentiation: An Introduction