12a: The Quotient Rule © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules.

12a: The Quotient Rule12a: The Quotient Rule

© Christine Crisp

““Teach A Level Maths”Teach A Level Maths”

Vol. 2: A2 Core Vol. 2: A2 Core ModulesModules

The Quotient Rule

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Module C3

OCR

The Quotient RuleThe following are examples of

quotients:

x

ey

x

1

(b)2

21

x

xy

(a)

(a) can be divided out to form a simple function as there is a single polynomial term in the denominator.

2

21

x

xy

For (b) we use the quotient rule.

2

2

2

1

x

x

xy

12 xy

32 xdx

dy

The Quotient RuleThe quotient rule gives us a way of

differentiating functions which are divided.

2vdx

dvu

dx

duv

dx

dy

v

uy

The rule is similar to the product rule.

where u and v are functions of x.Memory aid:

2bottom bottom grad x top – top grad x Bottom

The Quotient Rule

2vdxdv

udxdu

v

dx

dy

v

uy

2v

vuuv

dx

dy

v

uy

Or using the dash notation

2bottom bottom grad x top – top grad x Bottom

The Quotient Rule

1

2

x

xye.g. 1 Differentiate to find .

dx

dy

2vvu-uv

2v

dxdv

udxdu

v

dx

dy

v

uy

xdx

duu 2 1

dx

dvv

We now need to simplify.

v

uySolutio

n: 1 xvand

2

2

1

112

)()(

x

xxx

dx

dy

2xu

The Quotient Rule

2

2

)1(

)1(2

x

xxx

dx

dy

2

22

)1(

22

x

xxx

dx

dy

2

2

)1(

2

x

xx

dx

dy

2)1(

)2(

x

xx

dx

dy

We could simplify the numerator by taking out the common factor x, but it’s easier to multiply out the brackets. We don’t touch the denominator.

Now collect like terms:

and factorise:

We leave the brackets in the denominator as the factorised form is simpler.

Multiplying out numerator:

The Quotient RuleQuotients can always be turned into

products.

However, differentiation is usually more awkward if we do this.

21 x

e x

e.g. 12 )1( xe xcan be written

as

In the quotient above, and

xeu 21 xv

In the product , andxeu 12 )1( xv

( both simple functions )

( v needs the chain rule )

The Quotient RuleSUMMARY

2vdx

dvu

dx

duv

dx

dy

Otherwise use the quotient rule:If ,

v

uy

where u and v are both functions

of x

To differentiate a quotient:

Check if it is possible to divide out. If so, do it and differentiate each term.

The Quotient RuleExercis

eUse the quotient rule, where appropriate, to differentiate the following. Try to simplify your answers:

xe

xy

3

1.

2.

x

xy

2

2

3.

2

2

x

xy

The Quotient Rule

xe

xy

3

1.

23xdx

duu xe

dx

dvv

2vdxdv

udxdu

v

dx

dy

v

uy

and3xu xev

v

uy

2

323

)( x

xx

e

exex

dx

dy

2

2 3

)()(

x

x

e

xex

dx

dy xe

xx

dx

dy )(

32

Solution:

The Quotient Rule

2.

xdx

duu 2 1

dx

dvv

2vdx

dvu

dx

duv

dx

dy

v

uy

and2xu xv 2v

uy

2

2

2

122

)()()(

x

xxx

dx

dy

Solution:

x

xy

2

2

2

22

2

24

)( x

xxx

dx

dy

2

2

2

4

)( x

xx

22

4

)()(

x

xx

The Quotient Rule

22

2

x

x

xy

2

2

x

xy

234 xxdx

dy

Solution:

3.

2

2

x

xy

Divide out:

122 xxy

1

23

14

xx

The Quotient Rule

The Quotient 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 Quotient RuleThe following are examples of

quotients:

x

ey

x

1

(b)2

21

x

xy

(a)

(a) can be divided out to form a simple function as there is a single polynomial term in the denominator.

2

21

x

xy

For (b) we use the quotient rule.

2

2

2

1

x

x

xy

12 xy

32 xdx

dy

The Quotient Rule

2vdx

dvu

dx

duv

dx

dy

SUMMARY

Otherwise use the quotient rule:If ,

v

uy

where u and v are both functions

of x

To differentiate a quotient:

Check if it is possible to divide out. If so, do it and differentiate each term.

The Quotient Rule

12 xvxuv

uy and

Solution:

1

2

x

xye.g. 1 Differentiate to find .

dx

dy

2vdx

dvu

dx

duv

dx

dy

v

uy

xdx

du2 1

dx

dv

2

2

)1(

)1(2

x

xxx

dx

dy

We now need to simplify.

The Quotient Rule

2

2

)1(

)1(2

x

xxx

dx

dy

2

22

)1(

22

x

xxx

dx

dy

2

2

)1(

2

x

xx

dx

dy

2)1(

)2(

x

xx

dx

dy

We could simplify the numerator by taking out the common factor x, but it’s easier to multiply out the brackets. We don’t touch the denominator.

Now collect like terms:

and factorise:We leave the brackets in the denominator. ( A factorised form is considered to be simpler. )

Multiplying out numerator: