Differentiating : e x, lnx and the chain rule e x, lnx and the chain rule © Christine Crisp Objectives To understand why the differentials of lnx and ex.

Differentiating : Differentiating : eexx, lnx and the chain , lnx and the chain

rulerule

© Christine Crisp

Objectives

To understand why the differentials of lnx and ex are what they are!

To be able to differentiate more complex functions that are composite functions

),( 42Tangent at

2xy (2, 4)x

The Gradient at a point on a CurveDefinition: The gradient of a point on a curve

equals the gradient of the tangent at that point.e.g.

3

12

The gradient of the tangent at (2, 4) is

43

12 mSo, the gradient of the curve at (2, 4)

is 4

2xy

2m

The Rule for Differentiation

1. The gradient of this curve at any point is the tangent

2. The tangent changes at any one point for a curve

2xy

4m

2xy

6m

The gradient Dee-Y by Dee-XThe gradient for a function is defined as

h

xfxfLimit

)()(0

Use this to show that the gradient of y=x2 is 2x

xxf

xfx

2|

2

2 minutes for this !

Solution Like this....

02

2

2

2

)(0

2

222

22

asx

x

x

xxx

xxLim

Differentiate ex and ln x

From first principles

This means use the definition of gradient as a limit

1 minute to think about this

Let’s consider the derivative of the exponential function. Going back to our limit definition of the derivative:

h

eelime

dx

d xhx

0h

x

First rewrite the exponential using exponent rules.

h

eeelim

xhx

0h

Next, factor out ex. h

1eelim

hx

0h

Since ex does not contain h, wecan move it outside the limit.

h

xfhxflim

dx

d0h

xf

h

1elime

h

0h

x

Substituting h=0 in the limit expression results in the

indeterminate form , thus we will need to determine it.

00

We can look at the graph of and observe what happens as x gets close to 0. We can also create a table of values close to either side of 0 and see what number we are closing in on.

x1e

xfx

)(

x -.1 -.01 -.001 .001 .01 .1 y .95 .995 .999 1.0005 1.005 1.05

Graph

At x = 0, f(0) appears to be 1.

Table

As x approaches 0, y approaches 1.

X≠0 of course

We can safely say that from the last slide that 1

h

1elim

h

0h

Thus

Rule 1: Derivative of the Exponential Function

xxh

0h

xx e1eh

1elimee

dx

d

xx eedx

d

The derivative of the exponential function is the exponential function.

What about ln x

Method 1 – Using previous result

Method 2 – Consider Limits

What about ln x

Method 1 – Using previous result

Method 2 – Consider Limits

y = ln x

therefore ey = x ( raise both sides to e) dx/dy = ey. (Oh yes it does!!)

Since dx/dy * dy/dx = 1 1/(dx/dy) = dy/dx.

= 1/ey f'(x) = 1/x. (as ey=x)

y = ln x

f(x) = ln x f(x+h) = ln (x+h) f'(x) = lim(h->0) [ln (x+h) - ln x] / h

= lim(h->0) ln (x+h/x) / h

= lim(h->0) 1/h * ln(1+h/x) Since lim(h->0) ln(1+h/x) -> h/x where h ≠ 0,

f'(x) = 1/h * h/x = 1/x

Exercise

The Chain Rule

The gradient at a point on a curve is defined as the gradient of the tangent at that point.

The process of finding the gradient function is called differentiating.

The function that gives the gradient of a curve at any point is called the gradient function.

The rules we have developed for differentiating are:

A reminder of the differentiation done so far!

1 nn nxdx

dyxy

axax aedx

dyey

The Chain Rule

dx

dyxxy 168 36

We can find by multiplying out the brackets: dx

dy

However, the chain rule will get us to the answer without needing to do this ( essential if we had, for example, . )103 )4( x

)4(6 32 xx

2)( xxf Suppose and 4)( 3 xxg

))(( xgfyLet23 )4( x

25 246 xx

Differentiating a function of a function .

))(( xgf

)4( 3xf

The Chain Rule

2)( xxf Consider again and .4)( 3 xxg

Let 43 xuDifferentiating both these expressions:

2uy Then,

We must get the letters right.

23xd

d

ux

ud

d2

y

u

))(( xgfyLet 23 )4( x

The Chain Rule

2)( xxf Consider again and .4)( 3 xxg

Let 43 xuDifferentiating both these expressions:

2uy Then,

23xd

d

ux

ud

d2

y

u)4(2 3 x

))(( xgfyLet 23 )4( x

Now we can substitute for u

The Chain Rule

2)( xxf Consider again and .4)( 3 xxg

Let 43 xuDifferentiating both these expressions:

2uy Then,

23xd

d

ux

ud

d2

y

u)4(2 3 x

))(( xgfyLet 23 )4( x

Can you see how to get to the answer which we know is

?)4(6 32 xxdx

dy

We need to multiply by23x )4(2 3 x

The Chain Rule

23xdx

du u

du

dy2 )4(2 3 x

So, we have

So,

dx

du

du

dy

dx

dy

and to get we need to multiply by23x )4(2 3 xdx

dy

This expression is behaving like fractions with the ‘s on the r,h,s, cancelling.

du

The Chain Rule

23xdx

du u

du

dy2 )4(2 3 x

So, we have

So,

dx

du

du

dy

dx

dy

and to get we need to multiply by23x )4(2 3 xdx

dy

This expression is behaving like fractions with the s on the r,h,s, cancelling.

du

Although these are not fractions, they come from taking the limit of the gradient, which is a fraction.

The Chain Rule

Solution:

We need to recognise the function as and identify the inner function ( which is u ).

))(( xgf)(xg

5)(ye.g. 1 Find ifdx

dyx41

The Chain Rule5)(y

4dx

du

e.g. 1 Find if

dx

du

du

dy

dx

dy

dx

dy

45udu

dy

Solution:

x41

xu 41 Let Then 5uy

4)41(5 4 xdx

dy

4)41(20 x

Differentiating:

We need to recognise the function as and identify the inner function ( which is u ).

))(( xgf)(xg

We don’t multiply out the brackets

4)41(5 xSubstitute for u

Tidy up by writing the constant first

The Chain Rule

xy

52

1

e.g. 2 Find if

dx

dy

Solution: We can start in 2 ways. Can you spot them?

21

)52(

xyEither write

xu 52 and then let

Or, if you don’t notice this, start with xu 52

Thenu

y1

21

1

u

21

u

so 2

1uy

Always use fractions for indices, not decimals.

The Chain Rule

xy

52

1

e.g. 2 Find if

dx

dy

5dx

du

dx

du

du

dy

dx

dy

23

2

1 u

du

dy

xu 52

5)52(2

123

xdx

dy

23

)52(2

1 x

Solution: Whichever way we start we get

23

)52(2

5 x

21

uyand

The Chain Rule

The chain rule is used for differentiating functions of a function.

dx

du

du

dy

dx

dy

)(xgu where , the inner function.

If ,))(( xgfy

SUMMARY

The Chain RuleExercis

e

dx

dyUse the chain rule to find for the following:

32 )3( xy

1.

2.

8)21( xy

3.

43 xy4.

5.

51

1

xy

xy

31

2

Solutions:

1.

32 3 uyxu

xdx

du2

22 )3(6 xxdx

dy

222 )3(33 xudu

dy

dx

du

du

dy

dx

dy

The Chain RuleSolutio

ns2.

8)21( xy

3.

21

)43(43 xxy

821 uyxu

2dx

du 77 )21(88 xudu

dy

7)21(16 xdx

dy

21

43 uyxu

3dx

du21

21

)43(2

1

2

1 xu

du

dy

21

)43(2

3 x

dx

dy

dx

du

du

dy

dx

dy

dx

du

du

dy

dx

dy

The Chain RuleSolutions

4.

1)31(231

2

xx

y

1231 uyxu

3dx

du 22 )31(22 xudu

dy

2)31(6 xdx

dydx

du

du

dy

dx

dy

The Chain RuleSolutions5.

51

1

xy

4

2

11

5

xxdx

dy

511 uy

xu

22 1

xx

dx

du

44 1

155

xu

du

dy

dx

du

du

dy

dx

dy

11 x

The Chain Rule

The chain rule can also be used to differentiate functions involving e.

xedx

dy 22

e.g. 3. Differentiate xey 2

Solution: The inner function is the 1st operation on x, so u = 2x.

Let xu 2 uey

2dx

du xu eedu

dy 2

dx

du

du

dy

dx

dy

The Chain RuleExercis

eUse the chain rule to differentiate the

following: xey 3

1.

2.

xey

1

3.

2xey

Solutions:

1.

ueyxu 3

3dx

du

xedx

dy 33

xu eedu

dy 3

dx

du

du

dy

dx

dy

The Chain RuleSolutio

ns2.

xx

ee

y 1

3.

2xey

xu eedu

dy

ueyxu

1dx

du

ueyxu 2

xdx

du2

2xu eedu

dy

22 xxe

dx

dy

dx

du

du

dy

dx

dy

dx

du

du

dy

dx

dy

xedx

dy

The Chain Rule

Later we will want to reverse the chain rule to integrate some functions of a function.To prepare for this, we need to be able to use the chain rule without writing out all the steps.

32 )3( xye.g. For we know that 22 )3(6 xxdx

dy

has been multiplied by the derivative of

the outer function3)_____(

The derivative of the inner function which is

)3( 2 xx2

( I’ve put dashes here because we want to ignore the inner function at this stage. We must not differentiate it again. )

2)_____(3which is

The Chain Rule

So, the chain rule says

differentiate the inner function

multiply by the derivative of the outer function

e.g. dx

dyy

x4e

The Chain Rule

So, the chain rule says

differentiate the inner function

multiply by the derivative of the outer function

4e.g. dx

dyy

x4

( The inner function is ) x4

e

The Chain Rule

So, the chain rule says

differentiate the inner function

multiply by the derivative of the outer function

4 xe 4

( The outer function is )e

( The inner function is )

e.g. dx

dyy

x4e

x4

The Chain Rule

So, the chain rule says

differentiate the inner function

multiply by the derivative of the outer function

4

xe 44

xe 4

( The outer function is )

( The inner function is )

e.g. dx

dyy

x4e

x4e

The Chain RuleBelow are the exercises you have already done using the chain rule with exponential functions.

xey 3

1.

2.

xey

1

3.

2xey

See if you can get the answers directly.

Answers:

xedx

dy 33

1.

2.

xedx

dy

3.

22 xxe

dx

dy

Notice how the indices never change.

xe

The Chain Rule

TIP: When you are practising the chain rule, try to write down the answer before writing out the rule in full. With some functions you will find you can do this easily.However, be very careful. With some functions it’s easy to make a mistake, so in an exam don’t take chances. It’s probably worth writing out the rule.

The Chain Rule

The chain rule is used for differentiating functions of a function.

dx

du

du

dy

dx

dy

)(xgu where , the inner function.

If ,))(( xgfy

SUMMARY