5: The Chain Rule Christine Crisp Teach A Level Maths Vol. 2: A2 Core Modules

5: The Chain Rule5: The Chain Rule

© Christine Crisp

““Teach A Level Maths”Teach A Level Maths”Vol. 2: A2 Core Vol. 2: A2 Core

ModulesModules

The Chain Rule

Module C3

"Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages"

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 nxdxdyxy

axax aedxdyey

The Chain Rule

dxdyxxy 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 ))(( xgfyLet 23 )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.

23xdd

ux

udd 2

yu

))(( xgfyLet 23 )4( x

The Chain Rule



2uy Then,

23xdd

ux

udd 2

yu

)4(2 3 x

))(( xgfyLet 23 )4( x

Now we can substitute for u

The Chain Rule



2uy Then,

23xdd

ux

udd 2

yu

)4(2 3 x

))(( xgfyLet 23 )4( x

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

?)4(6 32 xxdxdy

We need to multiply by23x )4(2 3 x

The Chain Rule

23xdxdu

ududy 2 )4(2 3 x

So, we have

So,

dxdu

dudy

dxdy

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

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

du

The Chain Rule

23xdxdu

ududy 2 )4(2 3 x

So, we have

So,

dxdu

dudy

dxdy

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

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 ifdxdy x41

The Chain Rule5)(y

4dxdu

e.g. 1 Find if

dxdu

dudy

dxdy

dxdy

45ududy

Solution:

x41

xu 41 Let Then 5uy

4)41(5 4 xdxdy

4)41(20 x

Differentiating:


))(( 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

521

e.g. 2 Find ifdxdy

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

y 1

21

1

u 2

1u

so 2

1uy

Always use fractions for indices, not decimals.

The Chain Rule

xy

521

e.g. 2 Find ifdxdy

5dxdu

dxdu

dudy

dxdy

23

21

ududy

xu 52

5)52(21 2

3

xdxdy

23

)52(21

x

Solution: Whichever way we start we get

23

)52(25

x

21

uyand

The Chain Rule

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

dxdu

dudy

dxdy

)(xgu where , the inner function.

If ,))(( xgfy

SUMMARY

The Chain RuleExercis

e

dxdyUse the chain rule to find for the

following: 32 )3( xy

1.

2.

8)21( xy

3.

43 xy4.

5.

511

xy

xy

312

Solutions:

1.

32 3 uyxu

xdxdu 2

22 )3(6 xxdxdy

222 )3(33 xududy

dxdu

dudy

dxdy

The Chain RuleSolutio

ns2.

8)21( xy

3.

21)43(43 xxy

821 uyxu

2dxdu 77 )21(88 xu

dudy

7)21(16 xdxdy

21

43 uyxu

3dxdu

21

21

)43(21

21

xududy

21

)43(23

xdxdy

dxdu

dudy

dxdy

dxdu

dudy

dxdy

The Chain RuleSolutions

4.

1)31(2312

x

xy

1231 uyxu

3dxdu 22 )31(22 xu

dudy

2)31(6 xdxdydx

dududy

dxdy

The Chain RuleSolutions5.

511

xy

4

2115

xxdxdy

511 uyx

u

22 1

xx

dxdu

44 1155

xu

dudy

dxdu

dudy

dxdy

11 x

The Chain Rule

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

xedxdy 22

e.g. 3. Differentiate xey 2Solution: The inner function is the 1st operation on x, so u = 2x.

Let xu 2 uey

2dxdu xu ee

dudy 2

dxdu

dudy

dxdy

The Chain RuleExercis

eUse the chain rule to differentiate the

following: xey 3

1.

2.

xey 1

3.

2xey

Solutions:

1.

ueyxu 3

3dxdu

xedxdy 33

xu eedudy 3

dxdu

dudy

dxdy

The Chain RuleSolutio

ns2.

xx

ee

y 1

3.

2xey

xu eedudy

ueyxu

1dxdu

ueyxu 2

xdxdu 2

2xu eedudy

22 xxe

dxdy

dxdu

dudy

dxdy

dxdu

dudy

dxdy

xe

dxdy

The Chain RuleLater 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 xxdxdy

has been multiplied by the derivative of the outer function

3)_____(

The derivative of the inner function which is )3( 2 x

x2

( 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 RuleSo, the chain rule says

differentiate the inner function

multiply by the derivative of the outer functione.g.

dxdyy

x4e



multiply by the derivative of the outer function

4e.g. dxdyy

x4

( The inner function is ) x4

e




4 xe 4

( The outer function is )e

( The inner function is )

e.g. dxdyy

x4e

x4




4xe 44

xe 4

( The outer function is )


e.g. dxdyy

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:

xedxdy 33

1.

2.

xedxdy

3.

22 xxe

dxdy

Notice how the indices never change.

xe

The Chain RuleTIP: 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

The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied.For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.

The Chain Rule

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

dxdu

dudy

dxdy

)(xgu where , the inner function.

If ,))(( xgfy

SUMMARY

The Chain Rule5)(y

4dxdu

e.g. 1 Find if

dxdu

dudy

dxdy

dxdy

45ududy

Solution:

x41

xu 41 Let Then 5uy

4)41(5 4 xdxdy

4)41(20 x

Differentiating:


))(( xgf)(xg

We don’t multiply out the brackets

The Chain Rule

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

xedxdy 22

Let

e.g. Differentiate xey 2Solution: The inner function is the 1st operation on x so here it is 2x.

xu 2 uey

2dxdu xu ee

dudy 2

dxdu

dudy

dxdy

The Chain RuleLater 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 xxdxdy

has been multiplied by the derivative of the outer function

3)_____(

The derivative of the inner function which is )3( 2 x

x2

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

2)_____(3which is




4xe 44

xe 4

( The outer function is )


e.g. dxdyy

x4e

x4e

With exponential functions, the index never changes.

Documents

5: The Chain Rule Christine Crisp Teach A Level Maths Vol. 2: A2 Core Modules