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Differential Equations Worksheet
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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: [email protected]
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 [email protected]
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: [email protected]
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 [email protected]
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