Www.le.ac.uk Differentiation – Product, Quotient and Chain Rules Department of Mathematics University of Leicester

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Differentiation – Product, Quotient and Chain Rules

Department of MathematicsUniversity of Leicester

Content

Product Rule

Quotient Rule

Chain Rule

Inversion Rule

Introduction

Introduction• Previously, we differentiated simple

functions using the definition:

• Now, we introduce some rules that allow us to differentiate any complex function just by remembering the derivatives of the simple functions…

Next

h

xfhxf

dx

dfh

)()(lim 0

Product Quotient Chain InversionIntro

• The product rule is used for functions like:

where and are two functions.

• The product rule says:

• Differentiate the 1st term and times it by the 2nd, then differentiate the 2nd term and times it by the 1st.

Product rule

)()( xvxuy

dx

dvu

dx

duv

dx

dy

)(xu )(xv


Click here for a proof

Next

h

xvxuhxvxu )()()()( ......continue

h

hxvxuhxvhxu

h

xvxuhxvhxu

h

h

)()()()(lim

)()()()(lim

0

0dx

df

)()()( xvxuxf Let . Then:

Go back to Product Rule

dx

dvu

dx

duv

dx

dvxuhxv

dx

du

h

hxvxuhxu

h

h

)()(lim

)()()(lim

0

0

h

xvhxvxuh

)()()(lim

0

Go back to Product Rule

... DEQ

Product rule example

• Find .

Next

)ln( 2 xxdx

d

dx

dvu

dx

duv

dx

dy

2xu xu 2

xv

1xv ln

xxxx

xxx

ln21

)()2)((ln 2


Next

Differentiate these: dx

dvu

dx

duv

dx

uvd

)(

x

xx

x

ex

xee

xe

2

xx

xxxx

xxxx

22 cos2sin2

)2sin(cos2)2cos(sin2

)2sin(sin2)2cos(cos2

xx

x

x

x

xxx

ln93

9

3ln9

3

2

22

xxe

)2cos(sin xx

xx ln3 3


Check Answers Clear Answers Show Answers

Hint

Hint

Hint

• The quotient rule is used for functions like:

where and are two functions.

• The quotient rule says:

• This time, it’s a subtraction, and then you divide by .

Quotient rule

Next

)(xu )(xv)(

)(

xv

xuy

2vdxdvu

dxduv

dx

dy

2v



)()(

)()()()(lim

)()(

)()(

lim

0

0

xvhxhv

hxvxuxvhxu

hxvxu

hxvhxu

h

hdx

df

Let . Then:)(

)()(

xv

xuxf

Go back to Quotient Rule

......continue

)()(

)()()()()()(lim

)()(

)()()()()()()()(lim

0

0

xvhxhv

hxvxvxuxvxuhxu

xvhxhv

hxvxuxvxuxvxuxvhxu

h

h

......continue


2

0)(

)()()(

)()(lim

vdxdvu

dxduv

xuh

xvhxvxv

h

xuhxuh

)()(

1

xvhxv


... DEQ

Quotient rule example

• Find .

22 )23(

7

)23(

3646

xx

xx

Next

23

12

x

x

dx

d 12 xu 2u

3v23 xv

22 )23(

3)12(2)23(

x

xx

vdxdvu

dxduv

dx

dy


(give your answers as decimals)

Next

Differentiate these: 2v

dxdvu

dxduv

dxvu

d

1,1

2

xx

x

8,

2)2sin(

23)2sin(

xx

x

2,1

23

xx

x



Hint

Hint

Hint

• The chain rule is used for functions, , which have one expression inside another expression.

• Let be the inside part, so that now is just a function of .

• Then the chain rule says:

Chain rule

Next

dx

du

du

dy

dx

dy

, which has inside.2)2( xy 2x

2)( xxu 2uy , then

y

)(xuu

y



Instead of writing:

We write:

h

xfhxf

dx

dyh

)()(lim

0

The best way to prove the chain rule is to write the definition of derivative in a different way:

xa

xfaf

dx

dyxa

)()(lim

If we put , we see that these two definitions are the same.

xah

......continue

Go back to Chain Rule

xa

xuau

xuau

xufauf

xa

xufauf

xa

xa

)()(

)()(

))(())((lim

))(())((lim

We have .))(( xufy

dx

dy

......continue


dx

du

du

df

dx

du

ub

ufbf

xa

xuau

xuau

xufauf

ub

xaxa

)()(lim

)()(lim

)()(

))(())((lim

u(x) is just u, and u(a) is just a number, so we can call it b.

Then the first term matches the definition of .du

df


... DEQ

Chain rule example

• Find .))(sin( 2xdx

d

Next

)sin( 2xy 2xu

)cos(2cos2 2xxuxdx

du

du

dy

dx

dy

uy sin, so

udu

dycosx

dx

du2


differentiates to

Next

True or False?12xe 12 2

)1( xex

))12sin(( 2x ))12cos(()12(2 2 xx

3)1(sin x 2)1(sincos3 xx

differentiates to

differentiates to


True

True

True

False

False

False


Hint

Hint

Hint

Inversion Rule

• If you have a function that is written in

terms of y, eg.

Then you can use this fact:

• So if , then .

dydxdx

dy 1

)(yfdy

dx

)(

1

yfdx

dy

Next

132 yyx



First note that , because we’re

differentiating the function .

Then:

1dx

dx

dydxdx

dy 1

xxf )(

1dx

dx

dx

dy

dy

dx

This is a function, so we can divide by

it…We get: .

dx

du

du

dy

dx

dy

by the chain rule,

Go back to Inversion Rule

... DEQ

Inversion Rule Example

• A curve has an equation .

Find when .

• , therefore

• Then when ,

142 yyx

dx

dy

42 ydy

dx

42

1

ydx

dy

2

1

dx

dy

Next

1y

1y


2

2

12

2

dyxddx

yd

Note, it is NOT TRUE that

Next


Next

Find at the specified values of y:

dydxdx

dy 1

dx

dy

1,4

2

yyy

x

3,sin2

yyx



Hint

Hint

More complicated example

• Find .

• Quotient rule:

• Chain rule on :

– is ‘inside’, so let

– Then , so

1

12

2

x

x

e

e

dx

d

Next

12 xeu 12 xev

x2

1)( 2 xexu

wewuxxw )(,2)(

wewuxw )(,2)(xe

dx

dw

dw

du

dx

du 22


• Chain rule on also gives .

• Then quotient rule gives:

22

2222

22

2

)1(

2)1()1(2

1

1

x

xxxx

x

x

e

eeee

v

vuvu

e

e

dx

d

22

222222

)1(

2)(22)(2

x

xxxx

e

eeee

Next

1)( 2 xexv xedx

dv 22

22

2

)1(

4

x

x

e

e


Differentiate

xxxx

xxxx

cossin)cos(2

sin)sin(33

232

xxxxx cossin)sin(6 32

xxxx

xx

cossin)cos(2

sin)13cos(3

22

:sin)cos( 23 xxx

xxxx

xxxxx

cossin)cos(2

sin)cos()(3

233


Conclusion

• We can differentiate simple functions using the definition:

• We have found rules for differentiating products, quotients, compositions and functions written in terms of x.

• Using these two things we can now differentiate ANY function.

Next


h

xfhxf

dx

dfh

)()(lim 0

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Www.le.ac.uk Differentiation – Product, Quotient and Chain Rules Department of Mathematics University of Leicester