5: The Chain Rule Christine Crisp Teach A Level Maths Vol. 2: A2 Core Modules.
-
Upload
darrell-warner -
Category
Documents
-
view
237 -
download
4
description
Transcript of 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
2)( xxf Consider again and .4)( 3 xxg
Let 43 xuDifferentiating both these expressions:
2uy Then,
23xdd
ux
udd 2
yu
)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,
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:
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
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
The Chain RuleSo, the chain rule says
differentiate the inner function
multiply by the derivative of the outer function
4e.g. dxdyy
x4
( The inner function is ) x4
e
The Chain RuleSo, 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. dxdyy
x4e
x4
The Chain RuleSo, the chain rule says
differentiate the inner function
multiply by the derivative of the outer function
4xe 44
xe 4
( The outer function is )
( The inner 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:
We need to recognise the function as and identify the inner function ( which is u ).
))(( 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
The Chain RuleSo, the chain rule says
differentiate the inner function
multiply by the derivative of the outer function
4xe 44
xe 4
( The outer function is )
( The inner function is )
e.g. dxdyy
x4e
x4e
With exponential functions, the index never changes.