Study Advice Services

Part of the IFL Centre for Learning Development Mathematics Worksheet

Differentiation 1

This is one of a series of worksheets designed to help you increase your confidence in handling Mathematics. This worksheet contains both theory and exercises which cover:-

1. Simple functions of a variable 2. Product rule 3. Quotient rule 4. Function of a function (Chain rule)

There are often different ways of doing things in Mathematics and the methods suggested in the worksheets may not be the ones you were taught. If you are successful and happy with the methods you use it may not be necessary for you to change them. If you have problems or need help in any part of the work then there are a number of ways you can get help. For students at the University of Hull • Ask your lecturers • Contact the Study Advice Services on the ground floor of the Brynmor Jones

Library where you can access the Mathematics Tutor, or contact us by email. • Come to a Drop-In session organised for your department • Look at one of the many textbooks in the library. For others • Ask your lecturers • Access your Study Advice or Maths Help Service • Use any other facilities that may be available. If you do find anything you may think is incorrect (in the text or answers) or want further help please contact us by email.

Tel: 01482 466199 Web: www.hull.ac.uk/studyadvice Email: Studyadvice@hull.ac.uk

1

All formulas given can be proved and as they are in many textbooks, the proofs are not given here. This worksheet deals with products and quotients before the differentiation of a function of a function. 1. Indices

For all values of n ( 0)n ≠ , given nxy = then 1−= nnxdxdy

This means that Function Gradient Function Function Gradient Function

cy = (a constant)

0=dxdy

xy = 11 0 =×= xdxdy

2xy = xx

dxdy 22 1 ==

45xy =

33 2045 xxdxdy

=×=

4

32xxy +=

23

423 xx

dxdy

+=+= 11 −== xx

y 211 11

xx

dxdy

−=−= −−

Function Gradient Function 2

233 −== x

xy ( )

3312 6623

xxx

dxdy −

=−=−×= −−−

2

1

xxy == xx

xxdxdy

21

2

121

21

21

211

21 ==== −−

23

333

−−=−

= xx

y

( )52

9123

2

93 25

23

xxx

dxdy

==−= −−−−

23ty = ( ) ttdtdy 623 12 =×= −

233 ssr ==

2

321

23

231

23 sss

dsdr

=== −

243

243 −== tt

r

( )3

1243

232t

tddr

−=−= −−

θ

2

Examples

1. Find dxdy given

2

3 136x

xxy −+=

The expression can be written as

( ) ( )

3

332

12110

21

222

3

2

3

236236

21316

36

136136

xxxxx

xxxdxdy

xxxxx

xxx

xxxy

+−=+−=

−−−+×=

−+=

−+=−+

=

−−

−−−−

−−

giving

2. Work out ( )zzzdzd

+− 23 2

( )[ ]zfdzd is another way of saying differentiate ( )zf with respect to z .

( )zzzdzd

+− 23 2 = ( )21

23 2 zzzdzd

+−

= z

zzz2

126223 121 2

1

+−=

+−× −

3. Differentiate ( )yy 45 with respect to y

( )( )yy

dydzyy

dydz

yyyyyz

3245

245

29

44

27

27

29

21

5

555

==×=

==

or

= Let

(Giving the expression a ‘name’ such as z helps in the layout of the question. Note that the answer may be given in a number of different ways)

Exercise 1 A. Differentiate with respect to x

( )2

2

1.10

32.7

514.4

564.1

xxx

xx

xx

xx

+

+

+−

+−

( )( )

( )2

2

2

32

221

111

43.8

32

12.5

.2

xx

xx

xxx

xx

−

−+

+−

+

( )xx

xx

xx

xx

2

2

23

1.12

12.9

1.6

3137.3

−

+

+

−−

B. Find the following

( )( )( )

−+

−−

sss

dsd

ttdtd

2

2

31.3

273.1

( )( )( )[ ]321.4

11.22

+++

−

uuudud

yydyd

3

C. Differentiate with respect to y ( )( )

yyy

yy

1.4

121.12

2

++

−+

( )( )

yyy

yy35.5

113.2 33

++

+−

( )

112.6

1.32

3

+++

+

yyy

y

2. Product rule If you need to differentiate a function which is the product of two functions then, if it is not easy to multiply it out, you will need to use the product rule.

( )1365 34 −+= xxxy

is the product of the two functions 45x and ( )136 3 −+ xx but these terms are easy to multiply out.

( ) 45734 515301365 xxxxxxy −+=−+= ⇒ 346 2075210 xxxdxdy

−+=

( )( )23343 2524 +−−+= xxxxy is also a product which can be multiplied out but it is more complicated. There is a rule for differentiating such products. A “product” means that you have an expression which can be written as ( ) ( )y f x g x= ×

To differentiate the product function ( ) ( )xgxfy ×=

most formulas first write it as vuy ×= where ( )xfu = and ( )xgv =

then the rule states that dxdvuv

dxdu

dxdy

×+×= or '' uvvu +

or ( ) ( ) ( ) ( )xgxfxgxfdxdy '' ×+×=

(note: there are various forms of the ‘rule’ – keep using the one you remember as long as it gives the correct answer!) In the example above ( )( )23343 2524 +−−+= xxxxy

writing 23,343 2524 +−=−+= xxvxxu

we have xxvxxu 65',812' 43 −=+=

hence ( )( ) ( )( )xxxxxxxxuvvudxdy 6534323812'' 434253 −−+++−+=+=

This can be simplified if necessary.

4

Trigonometric Functions You should have met the differentials of the trigonometric functions xsin and .cos x These are used in example 2 below and the exercise following.

xdx

xdxdxdyxy

xdx

xdxdxdyxy

sin)(cos or sin= then cos If

cos)(sin or cos= then sin If

−=−=

==

Examples 1. Use the product rule to differentiate ( )( )21 42 −−= ssr

In this case

hence

Using

or, in this case,

( )( ) ( )( )sssssss

ssssdsdr

uvvudsdr

uvvudxdy

svsu

svsu

4464442

4122

''

''

4',2'

2,1

35535

324

3

42

++−=−++−=

−+−−=

+=

+=

=−=

−=−=

In practice it would have been easier to multiply the function out giving

( )( )

sssdsdr

sss

ssr

446

22

21

35

246

42

++−=

−++−=

−−=

giving

2. Find ( )( )φφφφ

cossin +− 1dd

Putting φφφ cos,sin +=−= 1vu then φφ sin',cos' −=−= vu 1

From the formula ( ) '' uvvuuvdd

+=φ

( )( )[ ] ( )( ) ( )( )

[ ]φφφφφ

φφφφ

φφφφφφφφφ

22

22

2

1

111

sincos-1 using sinsin

sinsincos

sinsincoscoscossin

2 =−=

+−−=

−−++−=+−dd

5

Note if you multiply the brackets out and then differentiate you get

( )( ) ( )φφφφφφφ

φφφφ

cossinsincoscossin −−+=+−dd

dd 1 which involves using the

product formula twice, on φφφφ cossin and cos . Exercise 2 Use the product rule to differentiate the following functions:

( )( )xx

xxx

cossin.

.

4

6431 223 −+

( )( )( )xx

xxx

cossin.

.

−+

−+

115

532 23

xx sin.3

3. Quotient Rule

To differentiate a function such as 231−+

=xxy requires another formula.

A quotient is a fraction containing variable terms (for instance terms in x ) in both the numerator and denominator which cannot be cancelled down or divided out. The rule for differentiating such functions is:-

To differentiate the quotient function ( )

)(xgxfy =

it is usual to write it as vuy = where ( )xfu = and ( )xgv =

then the rule states that 2v

dxdvuv

dxdu

dxdy ×−×

= or2

''v

uvvu −

In the example above

We have

giving

Applying the formula ( ) ( )

( )

( ) ( )22

22

235

233323

2331231''

3',1'23,1

231

−

−=

−

−−−=

−

×+−−=

−=

==−=+=

−+

=

xxxx

xxx

vuvvu

dxdy

vuxvxu

xxy

6

Examples

1 Differentiate the function 3

1322

2

−

+−=

xxxy

xvxuxvxxu 2',34'3,132 22 =−=⇒−=+−=

( )( ) ( )( )( )22

22

23

2132334''

−

+−−−−=

−=

x

xxxxxv

uvvudxdy

This simplifies to ( )22

2

3

9143

−

+−=

x

xxdxdy .

In some cases it may not be necessary to simplify the expression (such as when you need the value of the gradient at a particular point).

2 Differentiate the function t

tr sin=

1==⇒== ',cos',sin vtutvtu

2221

tttt

tttt

vuvvu

dxdy sincossincos'' −

=×−×

=−

=

Exercise 3 Differentiate the following functions (hint: in questions 7 & 8 you need to use product as well as quotient formulas)

( )( )

xxx

rrrr

xx

coscossin.

.

.

−

−++

+

17

2144

11

3

θθθ

sinsin.

.

cossin.

+

+

+

18

4

15

12

2s

xx

=

+

−+

yyy

xx

cossintan.

.

cossin.

9

16

13

ϕϕϕ

4. Function of a function (or Chain rule) Examples

1. Given ( )136 3 −+= xxy find .dxdy

This is a function (square root) of the function ( )136 3 −+ xx and cannot be simplified to get terms of the form .nx If you have a function which can be written in the form

( )[ ]xgfy = then it is a function ( )f of the function ( )g . Putting ( )xgu = the function can be written as ( )[ ] ).(ufxgfy == There is nothing special about the use of u as the dummy variable, it could be any convenient letter.

7

To differentiate the function of a function ( )[ ]xgfy =

first express it in the form )(ufy = where )(xgu = then the (chain) rule states

dxdu

dudy

dxdy

×= or ( ) ( )xgufdxdy '' ×=

In the above example above we have

( ) uxxy =−+= 136 3 where 136 3 −+= xxu

318136

21

23

21 2

121

+=⇒−+=

==⇒== −

xdxduxxu

uu

dudyuuy

Hence dxdu

dudy

dxdy

×= ( )1362

3183182

13

22

−+

+=+×=

xx

xxu

2. Differentiate the following: (i) xy 3sin= (ii) xy 3sin= (iii) θ33cos=r

(i) Numerically, for xy 3sin= you would work out 3x first and then take the sin of the answer; so xy 3sin= is a function (sin ) of a function (3x ).

,sinuy = where xu 3= ⇒ ,cosududy

= 3=dxdu

Hence xudxdu

dudy

dxdy 333 coscos =×=×=

(ii) Numerically, for ( )33 xxy sinsin == you would work out xsin first and then cube

the result; so xy 3sin= is a function (cube) of a function ( xsin )

,3uy = where xu sin= ⇒ ,3 2ududy

= xdxdu cos=

Hence xxxudxdu

dudy

dxdy cossincos 22 33 ==×=

(iii) Numerically, for ( )44 33 θθ coscos ==r you would work out θ3 first, take the

cosine of the value and finally cube the result; so θ34cos=r is a function of a function of a function and can be written as ,3cr = where uc cos= and θ3=u .

r is a function (cube) of a function (cos ) of a function (3θ)!

Extending the rule we can write θθ d

dududc

dcdr

ddr

××=

8

From θ34 === uuccr ,cos,

We have 34 3 =−==θd

duududcc

dcdr ,sin,

giving ( ) θθθ

331234 33 sincossin −=×−×= ucddr

3. Differentiate ( )( )654 233 +−= xxy This is a product: take 34 12'33 xuxu =⇒−=

and ( )65 2+= xv to find 'v requires the chain rule

write 6tv = where ( ) ( )554455 23056'2 +=×=×==⇒+= xxxtdxdt

dtdv

dxdvvxt

using '' uvvudxdy

+= gives ( )( ) ( ) ( ) 5544653 23033212 +×−++= xxxxxdxdy

which can be simplified further IF NECESSARY! It is highly unlikely that you’d meet anything as bad as this and doubtful if you’d have to even think of simplifying further!

Which comes first? To help decide which comes first, think of how you would calculate the value of the function if you were given a numerical value for the variable. If you had the function

( )xy sinln= , you would find the value of sinx first and then take the ln of the result. Hence in using the chain rule to differentiate the function you put sinu x= . The table below gives some further examples.

Function x6sin 12 +x x4cos

( )42 53 −+ xx 43 +xe

First (u) x6 12 +x xcos 53 2 −+ xx 3 4x + Second sin root power power exp 4. The volume of a mass at a time t seconds is given by ttttv cossin 22 += where v is measured in cubic metres. Find the rate of change of the volume after 4

π seconds and the first positive value of t for which the rate of change is zero.

Given ttttv cossin 22 += we need to find dtdv

both tttt cos and sin 22 are products so we need to use the product rule on each. write ttw sin2= as the product fg where 2f t= and sing t=

then ttttfggfdtdw cossin'' 22 +=+=

9

write ttu cos2= as the product hk where 2h t= and cosk t=

then ttthkkhdtdu sincos'' 22 −=+=

( ) ttttttttttttdtdu

dtdw

dtdv coscoscossincoscossin 22222 222 +=+=−++=+=

the rate of change of the volume, when t = 4π is ( ) ( ) 8512 4

24 .cos ≈

+ ππ

The rate of change of v is given by dtdv and it will be zero when

( ) 022 =+ tt cos ie when cos 0t = or 2 2 0t + =

cos 0t = when 32 2, ,t π π= …

2 2 0t + = has no real solution for t

hence first positive value is 2π=t

Exercise 4 Differentiate with respect to x

( )

( )( )127

3234

51

2

323

8

+

+−+

+

x

xxx

x

cos.

.

.

( )

( )128

415

12

22

52

+

+

++

x

x

xx

cos.

.

.

( )xx

x

x

2

3

9

16

73

sin.

sin.

sin.

+

Differentiate

θ4313

5103sin.

sin. u

( )xx

s

cossin.

cos.2

2

14

11 ( )23

2

3515

12

θcos.

cos. w

10

ANSWERS Exercise 1

2

22

3.7

144.4

68.1.

x

xx

x

−

+=+

−

−

A

3823.8

342

334.5

1.2

+−

++−

+

xx

xx

x

32

3

2

24.9

2

12

1.6

621.3

xx

xx

xx

−−

−

−

xxxxxx 2

25

23

23

21

23

21.10 −−=−−

−− 322232 22.11

xxxx −=− −−

xxxxxx

2323

25

23

2512 2

527

−=−−−.

76.1. −tB 3221.2yy

+− ss

213.3 2 −−− 11123.4 2 ++ uu

2

2

11.4

146.1

y

yy

−

−+C.

2

25

32

3.5

618.2

yy

yy

−

+

1.6

363.2 2 ++ yy

Exercise 2

( ) ( ) xxxxxx

xxxxxxxxxx

sinsincoscos.sincos.

cossin...

++−−

+−+

++−−

1154

32

51572241272301

22

23234

Exercise 3

( )211.1x+

xcos

.+1

12 ( )21

3ϕ

ϕϕ

cossin.

−

− ( ) ( )22

234

218102.4

−+

−−−

rrrrr

( )22 4

2.5+

−

s

s ( )212

1.6+

−

xxx

( ) xxx

x

xxcos

coscoscos

coscos.−

−−=

−

−+−1

1

1

2172

2

32

( )22

18

θ

θθθθ

sinsinsincos.

+

++ yy

22

19 seccos

. =

Exercise 4

( )758.1 +x

( )( )421215.2 xxx +++ x773 cos.

( )( )2232 3232633.4 +−+−+ xxxxx

11

x412.5+

( )2136 xx sincos. + ( ) ( ) ( )1188147 222 ++−+− xxxxx sincos.sin.

xxxx 229 sincossin. +

u

u2

5510 cos. ( )2211 ss sin. −

wwcossin. 212 − θθ 443613 2sincos.

xxx 32214 sincossin. − ( ) ( )222 339015 θθθ cossin. −

We would appreciate your comments on this worksheet, especially if you’ve found any errors, so that we can improve it for future use. Please contact the Maths tutor by email at Studyadvice@hull.ac.uk

updated 29th November 2004

The information in this leaflet can be made available in an alternative format on request from Sue Hodgson,

telephone 01482 466199

Study Advice Services

Part of the IFL Centre for Learning Development

Mathematics Worksheet

Differentiation 2

This is one of a series of worksheets designed to help you increase your confidence in handling Mathematics. This worksheet contains both theory and exercises which cover:-

1. Exponential functions 2. Logarithmic functions 3. Implicit Differentiation 4. Logarithmic Differentiation 5. Parametric Equations

There are often different ways of doing things in Mathematics and the methods suggested in the worksheets may not be the ones you were taught. If you are successful and happy with the methods you use it may not be necessary for you to change them. If you have problems or need help in any part of the work then there are a number of ways you can get help. For students at the University of Hull • Ask your lecturers • Contact the Study Advice Services on the ground floor of the Brynmor Jones

Library where you can access the Mathematics Tutor, or contact us by email. • Come to a Drop-In session organised for your department • Look at one of the many textbooks in the library. For others • Ask your lecturers • Access your Study Advice or Maths Help Service • Use any other facilities that may be available. If you do find anything you may think is incorrect (in the text or answers) or want further help please contact us by email.

Tel: 01482 466199 Web: www.hull.ac.uk/studyadvice Email: Studyadvice@hull.ac.uk

1

Some proofs are given in this worksheet - others may be found in a number of mathematics textbooks. If you have problems please ask. 1. Exponential Functions

It can be shown that xe is the function such that ( ) xx

edxed

=

Examples

Differentiate the following (i) xey 2= (ii) )(xfey = (iii) xe

xy2

1+=

(i) xey 2= is a function of a function

Writing 22 ==⇒==dxdue

dudyxuey uu and where

giving xu eedxdu

dudy

dxdy 222 =×=×=

(ii) ( )xfey = is a function of a function

Writing ( ) ( )xfdxdue

dudyxfuey uu '==⇒== and where

giving ( ) ( ) ( )xfu exfxfedxdu

dudy

dxdy '' =×=×=

Remembering ( ) ( ) ( )xf

xfexf

dxed '

)(=

will enable you to differentiate most exponential functions!

(iii) xexy 2

1+= is a quotient

writing xevxu 2,1 =+= gives xevu 22',1' ==

( )( )

( ) ( )xx

x

x

xx

ex

eex

e

exedxdy

vuvvu

dxdy

24

2

22

22

2

21221

211

''

+−=

−−=

×+−×=

−=

gives

using

Exercise 1 Differentiate the following

x

x

ex

e

2

7

5

1

.

.

316

2

x

e

ex

x

+.

. cos

( )x

x

xx

e

e

eee

17

32322

−

−+

.

.

( )

xe

ex

xx

2

8

4

.

. cossin +

2

2. Logarithmic functions The inverse function of xe is xelog which is usually written as xln (shorthand for natural or Napierian logarithms after Napier who developed them). For more information see the logs booklet. Given xy ln= then, from the definition of logarithms,

yex = which gives yedydx

= ⇒ xedx

dyy

11==

hence ( )

xdxxd 1

=ln

Extending this to differentiate ( )[ ]xfy ln= which is a function ( ln ) of the function ( )f x .

write ( )xfuuy == whereln ( )xfdxdu

ududy '1

==⇒ and

using the chain rule dxdu

dudy

dxdy

×=

giving ( ) ( ) ( )( )xfxfxf

xfdxdy ''1

=×=

Another important result to learn ( )[ ] ( )( )xf

xfxfdxd ')(ln =

Examples Differentiate the following functions

(i) ( )65 2 −= xy ln (ii)

++

=32

xxy ln (iii)

=

xxy

cossinln

2

(i) Using the above

( )65 2 −= xy ln gives 65

102 −

=x

xdxdy

(ii) Simplifying the expression gives

( ) )ln(lnln 3232

+−+=

++

= xxxxy

( ) ( )( )( ) ( )( )32

13223

31

21

++=

+++−+

=

+−

+=

xxxxxx

xxdxdy Hence

Note you could do this without simplifying but it is more difficult!

3

(iii) Simplifying the expression gives

( ) ( )

( ) ( )

( )112

2

2

22222

22

=++

=+

=

−−=

−=

−=

=

xxxxx

xxxx

xx

xx

dxdy

xx

xxxxy

sincos usingcossin

coscossin

sincoscossin

sincos

coslnsinln

coslnsinlncossinln

Exercise 2 Differentiate the following

( )

−+

xxx

xx

x

cossinln.

ln.

ln.

7

124

71

2

( )

+

−

−

−

118

3255

62

2

xx

xx

x

ln.

ln.

ln.

( )

+

+

+

+

2

3

2

2

2549

16

33

xx

xx

x

ln.

ln.

ln.

3. Implicit Functions A function such as 53 35 +−+= xxxy is called an explicit function as y is explicitly given in terms of x . A function such as 15353 235 =−+−+ xyxyyxx is called an implicit function as y is not given explicitly in terms of x nor x in terms of y . An implicit function can be differentiated with respect to x as it stands. Consider 1533 22 =−+−+ xxyyyx Differentiating each term with respect to x we get:

( ) ( )dx

ddx

xddxxyd

dxyd

dxyd

dxxd 15)3()()(3)( 22

=−+−+

To differentiate a function of y with respect to x we need

to use the chain rule ( )[ ] ( )[ ]dxdy

dyyfd

dxyfd

×=

giving ( ) ( )dxdy

dxdy

dyyd

dxyd 333

=×= and dxdyy

dxdy

dyyd

dxyd 2)()( 22

=×=

4

using the product formula dxdyxy

dxdyxy

dxxd

dxxyd

+=×+×=11

)()(

Putting these together we have: ( ) ( ) ( ) ( )

( )

xyxy

dxdy

xydxdyxy

dxdyxy

dxdyy

dxdyx

dxd

dxxd

dxxyd

dxyd

dxyd

dxxd

+−−−

=

−−=+−

=−++−+

=−+−+

2323

2323

03232

1533 22 )()(

Example Find the gradient of the curve 1035222 =+−++ yxxyyx at the points where 1x = First we need to find the values of y when 1x = Putting 1x = we get 0145103521 22 =−+⇒=+−++ yyyyy which gives ( )( ) 72072 −==⇒=+− yyyy or notice that there are two points to consider (1, 2) and (1, -7) Differentiating the function 1035222 =+−++ yxxyyx

gives ( ) ( ) ( ) ( )dx

ddx

yddx

xddx

xyddxyd

dxxd 1035222

=+−++)()(

0352222 =+−+++dxdy

dxdyxy

dxdyyx

giving 322

225++

−−=

xyyx

dxdy

5 2 4 1at (1, 2) 4 2 3 95 2 14 17at (1, 7) 14 2 3 9

dyPdxdyQdx

− −= = = −

+ +− +

= − = = −− + +

Note you could substitute in and find

the value of dxdy

without making it the

subject. The sketch of the graph shows the two points P and Q. From the sketch you can see that the gradient is negative in both cases.

y

x 1

P

Q

5

Exercise 3

1. In the following find dxdy in terms of x and y

(i) 1022 =+ yx (ii) yxyx 710322 22 +=++ (iii) 6322 =+− xyyx (iv) 032 323 =−+ yxyx

2. Find the gradient of the curve 106 22 =+ yx at the points where 2x = . 3. Find the gradient of the curve 154 23 +=+ yxyx at the points where 2x = . 4. Logarithmic Differentiation The function xay = cannot be differentiated by any of the methods developed so far. But taking the natural logarithm of both sides overcomes the problem! To solve xay =

take logs ( ) axay x lnlnln ==

differentiate ( ) ( )dx

axddx

yd lnln=

By the chain rule the left hand side gives ( ) ( )dxdy

ydxdy

dyyd

dxyd 1

==lnln

the right hand side gives ( ) ( ) adx

xdadx

axd lnlnln==

putting these together gives adxdy

yln=

1

hence ( ) ( ) xaayadxdy lnln ==

This method can simplify differentiation in a number of cases, as shown in the following examples. Examples (The first two could be differentiated as quotients.)

1. Find dxdy given the function

xxy

cossin

= (ie tanx)

Taking logs gives xxy coslnsinlnln −=

seccoscos

sincossincossin

cossincossinsincos

cossin

sincos ateDifferenti

xxx

xxx

yxxdx

dyxxxx

xxxx

xx

dxdy

y

22

22

111

11

==×=×=

=+

=−

−=

6

The result should be known xxdxd 2sec)(tan =

2. Find dxdy given the function ( ) xx

xxycos

sin1+

=

( ) ( ) ( ) ( )[ ]

( ) ( ) ( ) ( )xxxx

xxxxxx

xxy

coslnlnsinlnln

coslnsinlncos

sinlnln

−+−+=

+−=

+

=

1

11

Differentiating gives

( )

+

+−+

+=

+

+−+=

++

−+=

xx

xxx

xxxxx

xx

xxx

xy

dxdy

xx

xxx

xdxdy

y

cossin

sincos

cossin

cossin

sincos

cossin

sincos

111

1

1111

111

which is a lot easier than using the quotient method. It could be ‘simplified’ but this rarely needs to be done.

3. Find dxdy given the function xxy =

Take natural logs ( ) ( ) ( )xxxy x lnlnln ==

Differentiate ( ) ( )xxx

xdxdy

ylnln +=+×= 111 (using the product rule)

( )[ ] ( )[ ]xxxydxdy x lnln +=+= 11

Exercise 4 Use logarithmic differentiation to differentiate the following:

( )( )

( ) ( )432

17

4

21

2 −+

+=

=

=

xx

xy

uv

r

u

.

sin.

. θ

( )1

1.5

.2

+

+=

=

xexey

xy

x

x

x

( )

xxy

tts tt

cossin.

sinsin.

+=

==

16

32

7

5. Parametric Differentiation When a function is given in parametric form it means that x and y are given in terms of another variable, the parameter. i.e. ( )x f t= , ( )y g t= .

tytx 2 ,2 == are parametric equations. Frequently the parameter can be eliminated.

( )parabola a of equation the , or

hence but

xy

yyxtx

ytty

4

2

2

2412

212

21

=

===

=⇒=

To find the gradient of such a function in parametric form we need to use the chain

rule '' or as writtenbe can which

xy

dtdxdtdy

dxdy

dxdt

dtdy

dxdy

==

ttdxdy

dtdyt

dtdx

tytx

122

22

22

==

==

==

hence

, have we

, Given

In this case we can also find the gradient using the Cartesian equations:

yydxdydxdyy

xy

224

4

4

==

=

hence

=2 have we

Given 2

Comparing the two answers, as 2 12 then y ty t

= = so the two answers are the same

(as expected!) Examples 1. Find the gradient of the curve given by 32 π=== ttytx whencos,sin .

Finddtdy and

dtdx

and use

dtdx

dtdy

dxdy

÷= =''

xy

Finally substitute for t

( )( ) 32

22

22

22

2

21

23

3

32

3 −=

−

=−

==

−==

−==

==

π

ππ

cos

sin, when

cossin

''

sin,cos

cos,sin

dxdyt

tt

xy

dxdy

tdtdyt

dtdx

tytx

8

Notes a) It would be possible to eliminate t and obtain the Cartesian equation 221 xy −= which will give the same value for the gradient.

b) By putting ,cossinsin xxx 22 = t

tdxdy

cossin22−

= can be simplified to

sincos

cossin tt

ttdxdy 44

−=−

= if necessary.

2. Find the gradient of the curve given by cos,sin θθθ −=+= 1yx when 2

πθ = and when πθ = .

θθ

θθθ

θθ sin cos;cossin =⇒−=+=⇒+=ddyy

ddxx 11

θθ

cossin

''

+==

1xy

dxdy

when ( )( ) ,

cos

sin1

011

1 2

22 =

+=

+==

π

ππθ

dxdy tangent at 45o

when, ,cos

sin∞=

−=

+==

110

1 πππθ

dxdy tangent vertical

(really the value is indeterminate)

Notes a) θθ sin+=x and θcos−= 1y cannot be made into a simple Cartesian

equation!

b)θ

θcos

sin+

=1dx

dy can be simplified by putting ( ) ( )222 θθθ cossinsin =

and ( ) 12 22 −= θθ coscos giving

( ) ( )( )

( ) ( )( ) ( )22

222

22

22

2

2

121

2θ

θ

θθ

θ

θθtan

cos

cossin

cos

cossin==

−+=

dxdy

Exercise 5 In questions 1 to 8 find dxdy in terms of the given parameter.

11,1.7

1,.5

1,41.3

3,31.12

2

−+

==

+==

+=−=

−=+=

ssy

sx

yex

sysx

tytx

θθ

tty

tx

eeyeex

yxttyttx

uuuu

+=

+=

−=+=

+==−=+=

−−

1118

6

124232

2

22

,.

,.

cos,sin.,.

φφ

9

In questions 9 to 14 find the gradient of the curve at the given point.

1113

211

223319

3

2

=+==

===

=+=−=

sssysx

yx

ttytx

);ln(,ln.

;sin,cos.

;,.πααα

0214

011122210

2

22

=−=+=

=+=−=−=−=−=

rreyeex

yxtttyttx

rr ;,.

;cos,sin.;,.

φφφ

ANSWERS

Exercise 1

( ) ( )2

2

42

27

1281733625

46432712

xxe

ee

xexeexx

exxxeexeex

xx

xxx

xxxxxx

−=

−−+

−+−

−

+

....

)sin(cos..sin.. )cos(sincos

Exercise 2

xxxxxx

xxx

xx

xxxxxxx

xx

xx1117

12

1126

323

32215

12

2243

23121122

−=−+++

=+

−

−−

=−

−−+

+

+−

sincostancot.

)(.

)(.....

8. Simplify to )ln()ln( 112 21 +−− xx ; answer ( ) ( )( )112

5312

11

2+−

+=

+−

− xxx

xx

9. Hint first simplify as above; answer ( )

( )( ) ( )

4 5 72 104 3 5 2 3 4 5 2

xx x x x

−− =

+ + + +

Exercise 3

1. (i) yx

− (ii) yx

4743

−+ (iii)

xyxy

3223

−+ (iv) ( )xyy

yx

xyy

yx2

2

63

36 22

2

22

−+

=−

+

2. 12 1 grad6 3

dy xx ydx y

−= ⇒ = ± = = ±

3. 23 4 8 202 1, 7 grad ,

2 4 3 3dy x yx ydx y x

+= ⇒ = = = −

−

Exercise 4

( ) ( ) ( ) ( ) ( )( )

( ) ( ) ( )22

2

2 324

124761

25

14312221

+−

++−

+

−−

+

+=+

xx

xxxxe

xe

uuutttttsxx

x

x

uutx

.sin..

lncos.sinsincossinln.ln.ln. θ

10

Exercise 5

38.9

.5

32.1

−

−

−θe

t

56.10

.6

3222.2

−

−

+

+−

−

−

uu

uu

eeee

tt

( )

32.11

1

2.7

2.3

2

2

−

−

ss

s

( )

21142

2113012

28

4224

2

.ln..

.

sincos

sin.

+

+−

−=−

tt

φφφ

We would appreciate your comments on this worksheet, especially if you’ve found any errors, so that we can improve it for future use. Please contact the Maths tutor by email at Studyadvice@hull.ac.uk

Updated 29th November 2004

The information in this leaflet can be made available in an alternative format on request from Sue Hodgson,

telephone 01482 466199