12: The Quotient Rule12: 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

AQA

Edexcel

MEI/OCR

The Quotient RuleThe following are examples of

quotients:

x

xy

sin(a)

x

ey

x

1

(b)

2

21

x

xy

(c)

x

xy

cos

sin(d)

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

2

21

x

xy

For the others 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.

This rule can be derived from the product rule but it is complicated. If you want to go straight to the examples, click on the box below.

where u and v are functions of x.

Examples

The Quotient RuleWe can develop the quotient rule by using the

product rule!1 vuy

The problem now is that this v is not the same as the v of the product rule. That v is replaced by .

1v

So, becomesdx

dvu

dx

duv

dx

dy

dx

vdu

dx

duv

dx

dy )( 11

dx

vdu

dx

du

vdx

dy )(1 1

Simplifying

dx

vd )( 1Part of the 2nd term, , is the derivative of

but with respect to x not v.

1v

vuy

v

uy

1

The Quotient Rule

dx

dv

v

2

1

dx

dv

v

u

dx

du

v

v

dx

dy22

2vdx

dvu

dx

duv

dx

dy

dx

dv

dv

vd

dx

vd

)()( 11

We use the chain rule:

dx

vd )( 1So,

dx

dv

v

u

dx

du

vdx

dy2

1

Make the denominators the same by multiplying the numerator and denominator of the 1st term by v.

Write with a common denominator:

dx

vdu

dx

du

vdx

dy )(1 1Then,

The Quotient Rule

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.

2xuv

uy Solutio

n: and 1xv

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.

3.

x

xy

2

2

4

cos

x

xy

4.

2

2

x

xy

The Quotient Rule

xe

xy

3

1.

23xdx

du xe

dx

dv

2vdx

dvu

dx

duv

dx

dy

v

uy

and3xu xev

v

uy

2323

x

xx

e

exex

dx

dy

22 )3(

x

x

e

xex

dx

dy

xe

xx

dx

dy )3(2

Solution:

The Quotient Rule

4

cos

x

xy

2.

xdx

dusin 34x

dx

dv

2vdx

dvu

dx

duv

dx

dy

v

uy

andxu cos 4xv v

uy

24

34 cos4sin

x

xxxx

dx

dy

8

3 )cos4sin(

x

xxxx

dx

dy

Solution:

5x5

)cos4sin(

x

xxx

The Quotient Rule

3.

xdx

du2 1

dx

dv

2vdx

dvu

dx

duv

dx

dy

v

uy

and2xu xv 2v

uy

2

2

)2(

)1()2(2

x

xxx

dx

dy

Solution:

x

xy

2

2

2

22

)2(

24

x

xxx

dx

dy

2

2

)2(

4

x

xx

2)2(

)4(

x

xx

The Quotient Rule

22

2

x

x

xy

2

2

x

xy

234 xxdx

dy

Solution:

4.

2

2

x

xy

Divide out:

122 xxy

1

23

14

xx

The Quotient Rule

))(cos(cos

)sin)((sin))(cos(cos

xx

xxxx

dx

dy

v

uy xu sin xv cos

xdx

ducos x

dx

dvsin

x

xx2

22

cos

sincos

We can now differentiate the trig function

x

xxy

cos

sintan

by writing

xy tan

The Quotient Rule

1sincos 22 xx

x

xx

dx

dy2

22

cos

sincos xy tanSo,

This answer can be simplified:

xcos

1 is defined as

xsecAlso,

xdx

dy 2sec xy tanSo,

xdx

dy2cos

1

The Quotient Rule

Those of you taking the OCR/MEI spec can omit the next ( final exercise ).

The Quotient RuleExercis

eUse the quotient rule ( or, for (a) and (b), the chain rule ) to find the derivatives with respect to x of

xy cosec(a) xy sec(b) xy cot(c)Before you check the solutions, look in your formula books to see the forms used for the answers. Try to get your answers into these forms.

The Quotient Rule

(a) Solution:

xxysin

1cosec

2)(sin

)(cos1)0(sin

x

xx

dx

dy

xx

x

sinsin

cos

xx

x

sin

1

sin

cos

xx coseccot )cotcosec( xx

The Quotient Rule

xxycos

1sec

2)(cos

)sin(1)0(cos

x

xx

dx

dy

xx

x

coscos

sin

xx

x

cos

1

cos

sin

xx sectan )tansec( xx

(b) Solution:

The Quotient Rule

x

xxysin

coscot

2)(sin

coscos)sin(sin

x

xxxx

dx

dy

x

xx2

22

sin

)cos(sin

x2sin

1

x2cosec

(c) Solution:

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

xy

sin(a)

x

ey

x

1

(b)

2

21

x

xy

(c)

x

xy

cos

sin(d)

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

2

21

x

xy

For the others 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:

The Quotient Rule

))(cos(cos

)sin)((sin))(cos(cos

xx

xxxx

dx

dy

xy tanWe can now differentiate the trig function

x

xxy

cos

sintan

by writing

v

uy xu sin xv cos

xdx

ducos x

dx

dvsin

x

xx2

22

cos

sincos

The Quotient Rule

1sincos 22 xx

x

xx

dx

dy2

22

cos

sincos xy tanSo,

This answer can be simplified:

xcos

1 is defined as

xsecAlso,

xdx

dy 2sec xy tanSo,

xdx

dy2cos

1