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Murthi Katherason
MathematicsWorksheet Differentiation 2
This is one of a series of worksheets designed to help you increase your confidencein handling Mathematics. This worksheet contains both theory and exercises whichcover:-
1. Exponential functions 2. Logarithmic functions3. Implicit Differentiation 4. Logarithmic Differentiation5. Parametric Equations
There are often different ways of doing things in Mathematics and the methodssuggested in the worksheets may not be the ones you were taught. If you aresuccessful and happy with the methods you use it may not be necessary for you tochange them. If you have problems or need help in any part of the work then thereare a number of ways you can get help.
For students at the University of Hull Ask your lecturers Contact the Study Advice Service in 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 wantfurther help please contact us by email.
Email: [email protected]: 0193502214
1
Some proofs are given in this worksheet - others may be found in a number ofmathematics textbooks. If you have problems please ask.
1. Exponential Functions
It can be shown that xe is the function such that xx
edx
ed
Examples
Differentiate the following (i) xey 2 (ii) )(xfey (iii)xe
xy
2
1
(i) xey 2 is a function of a function
Writing 22 dx
due
du
dyxuey uu and where
giving xu eedx
du
du
dy
dx
dy 222
(ii) xfey is a function of a function
Writing xfdx
due
du
dyxfuey uu ' and where
giving xfu exfxfedx
du
du
dy
dx
dy''
Remembering xfxf
exfdx
ed'
)(
will enable you to differentiate mostexponential functions!
(iii)xe
xy
2
1 is a quotient
writing xevxu 2,1 gives xevu 22',1'
xx
x
x
xx
e
x
e
ex
e
exe
dx
dy
v
uvvu
dx
dy
24
2
22
22
2
21221
211
''
gives
using
Exercise 1Differentiate the following
x
x
ex
e
2
7
5
1
.
.
3
16
2
x
e
ex
x
.
. cos
x
x
xx
e
e
eee
17
32322
.
.
x
e
e
x
xx
2
8
4
.
. cossin
2
2. Logarithmic functionsThe inverse function of xe is xelog which is usually written as xln (shorthand fornatural or Napierian logarithms after Napier who developed them). For moreinformation see the logs booklet.
Given xy ln then, from the definition of logarithms,yex which gives ye
dy
dx
xedx
dyy
11
hence
xdx
xd 1
ln
Extending this to differentiate xfy ln which is a function ( ln ) of the function ( )f x .
write xfuuy whereln xfdx
du
udu
dy'
1 and
using the chain ruledx
du
du
dy
dx
dy
giving xf
xfxf
xfdx
dy ''
1
Another important result to learn xf
xfxf
dx
d ')(ln
ExamplesDifferentiate the following functions
(i) 65 2 xy ln (ii)
3
2
x
xy ln (iii)
x
xy
cossinln
2
(i) Using the above
65 2 xy ln gives65
102
x
x
dx
dy
(ii) Simplifying the expression gives
)ln(lnln 323
2
xxx
xy
32
1
32
233
1
2
1
xxxx
xxxxdx
dyHence
Note you could do this without simplifying but it is more difficult!
3
(iii) Simplifying the expression gives
112
2
2
22222
22
xxxx
x
xx
xx
x
x
x
x
dx
dy
xx
xxx
xy
sincosusingcossin
coscossin
sincoscos
sinsincos
coslnsinln
coslnsinlncossinln
Exercise 2Differentiate the following
xx
x
xx
x
cossinln.
ln.
ln.
7
124
71
2
1
18
32
55
62
2
x
x
x
x
x
ln.
ln.
ln.
2
3
2
2
25
49
16
33
x
x
x
x
x
ln.
ln.
ln.
3. Implicit FunctionsA function such as 53 35 xxxy is called an explicit function as y is explicitlygiven in terms of x .
A function such as 15353 235 xyxyyxx is called an implicit function as y isnot 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 xxyyyxDifferentiating each term with respect to x we get:
dx
d
dx
xd
dx
xyd
dx
yd
dx
yd
dx
xd 15)3()()(3)( 22
To differentiate a function of y with respect to x we need
to use the chain rule dx
dy
dy
yfd
dx
yfd
giving dx
dy
dx
dy
dy
yd
dx
yd3
33 and
dx
dyy
dx
dy
dy
yd
dx
yd2
)()( 22
4
using the product formuladx
dyxy
dx
dyxy
dx
xd
dx
xyd
11
)()(
Putting these together we have:
xy
xy
dx
dy
xydx
dyxy
dx
dyxy
dx
dyy
dx
dyx
dx
d
dx
xd
dx
xyd
dx
yd
dx
yd
dx
xd
23
23
2323
03232
1533 22 )()(
ExampleFind 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 ornotice that there are two points to consider (1, 2) and (1, -7)
Differentiating the function 1035222 yxxyyx
gives dx
d
dx
yd
dx
xd
dx
xyd
dx
yd
dx
xd 1035222
)()(
0352222 dx
dy
dx
dyxy
dx
dyyx
giving322
225
xy
yx
dx
dy
5 2 4 1at (1, 2)4 2 3 95 2 14 17at (1, 7)14 2 3 9
dyP
dxdy
Qdx
Note you could substitute in and find
the value ofdx
dywithout making it the
subject.
The sketch of the graph shows the two points P and Q. From the sketch you cansee that the gradient is negative in both cases.
y
x
1
P
Q
5
Exercise 3
1. In the following finddx
dy 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 DifferentiationThe 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
axd
dx
yd lnln
By the chain rule the left hand side gives dx
dy
ydx
dy
dy
yd
dx
yd 1
lnln
the right hand side gives a
dx
xda
dx
axd lnlnln
putting these together gives adx
dy
yln
1
hence xaayadx
dy lnln
This method can simplify differentiation in a number of cases, as shown in thefollowing examples.
Examples (The first two could be differentiated as quotients.)
1. Finddx
dy given the functionx
xy
cossin
(ie tanx)
Taking logs gives xxy coslnsinlnln
seccoscos
sincossincossin
cossincossinsincos
cossin
sincosateDifferenti
xxx
x
xxy
xxdx
dy
xxxx
xx
x
x
x
x
dx
dy
y
22
22
111
11
6
The result should be known xxdxd 2sec)(tan
2. Finddx
dy given the function xx
xxy
cossin1
xxxx
xxxxxx
xxy
coslnlnsinlnln
coslnsinlncos
sinlnln
1
11
Differentiating gives
x
x
xx
x
xxx
xx
x
x
xx
x
xy
dx
dy
x
x
xx
x
xdx
dy
y
cossin
sincos
cossin
cossin
sincos
cossin
sincos
1
11
1
1
11
1
111
which is a lot easier than using the quotient method. It could be ‘simplified’ but thisrarely needs to be done.
3. Finddx
dy given the function xxy
Take natural logs xxxy x lnlnln
Differentiate xxx
xdx
dy
ylnln 1
11 (using the product rule)
xxxydx
dy x lnln 11
Exercise 4Use logarithmic differentiation to differentiate the following:
432
17
4
21
2
xx
xy
uv
r
u
.
sin.
.
1
1.5
.2
xe
xey
xy
x
x
x
x
xy
tts tt
cossin.
sinsin.
16
32
7
5. Parametric DifferentiationWhen a function is given in parametric form it means that x and y are given in termsof another variable, the parameter. i.e. ( )x f t , ( )y g t .
tytx 2,2 are parametric equations. Frequently the parameter can be eliminated.
parabolaaofequationthe,or
hencebut
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''oras writtenbecan which
x
y
dt
dxdt
dy
dx
dy
dx
dt
dt
dy
dx
dy
ttdx
dydt
dyt
dt
dx
tytx
1
2
2
22
22
hence
,havewe
,Given
In this case we can also find the gradient using the Cartesian equations:
yydx
dydx
dyy
xy
2
2
4
4
4
hence
=2havewe
Given 2
Comparing the two answers, as 2 12 theny ty t
so the two answers are the same
(as expected!)
Examples1. Find the gradient of the curve given by
32 ttytx whencos,sin .
8
Finddt
dy anddt
dx
and use
dt
dx
dt
dy
dx
dy =
'
'
x
y
Finally substitute for t
3222
22
22
2
21
23
3
32
3
cos
sin,when
cossin
''
sin,cos
cos,sin
dx
dyt
t
t
x
y
dx
dy
tdt
dyt
dt
dx
tytx
Notes a) It would be possible to eliminate t and obtain the Cartesian equation221 xy which will give the same value for the gradient.
b) By putting ,cossinsin xxx 22 t
t
dx
dy
cossin 22
can be simplified to
sincos
cossint
t
tt
dx
dy4
4
if necessary.
2. Find the gradient of the curve given by cos,sin 1yx when2 and
when .
sincos;cossin d
dyy
d
dxx 11
cossin
''
1x
y
dx
dy
when
,cos
sin1
01
1
12
22
dx
dy tangent at 45o
when, ,cos
sin
11
0
1
dx
dy tangent vertical
(really the value is indeterminate)
Notes a) sinx and cos 1y cannot be made into a simple Cartesian
equation!
b)
cos
sin
1dx
dy can be simplified by putting 22
2 cossinsin
and 122
2 coscos giving
2
22
22
22
22
2
2
121
2
tancos
cossin
cos
cossin
dx
dy
Exercise 5 In questions 1 to 8 finddx
dy in terms of the given parameter.
9
1
1,
1.7
1,.5
1,41.3
3,31.12
2
s
sy
sx
yex
sysx
tytx
t
ty
tx
eeyeex
yx
ttyttx
uuuu
11
18
6
124
232
2
22
,.
,.
cos,sin.,.
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.
;,.
0214
01112
2210
2
22
rreyeex
yx
tttyttx
rr ;,.
;cos,sin.;,.
ANSWERS
Exercise 1
2
2
42
27
128
17
33625
46432712
x
xe
ee
x
exeexx
exxxeexeex
xx
xxx
xxxxxx
....
)sin(cos..sin.. )cos(sincos
Exercise 2
xxxxxx
xx
x
xx
xxxxxx
x
x
x
xx
1117
1
2
1
126
32
3
32
215
12
224
3
23
12
11
22
sincostancot.
)(.
)(.....
8. Simplify to )ln()ln( 11221 xx ; answer 112
53
12
1
1
2
xx
x
xx
9. Hint first simplify as above; answer
4 5 72 104 3 5 2 3 4 5 2
x
x x x x
Exercise 3
1. (i)y
x (ii)
y
x
47
43
(iii)
xy
xy
32
23
(iv) xyy
yx
xyy
yx
2
2
63
36 22
2
22
2. 12 1 grad6 3
dy xx y
dx y
10
3.23 4 8 202 1, 7 grad ,
2 4 3 3dy x y
x ydx y x
Exercise 4
22
2
2 324
12476
1
25
14312221
xx
xxx
xe
xe
uuutt
tttsxx
x
x
uutx
.sin..
lncos.sinsincossinln.ln.ln.
Exercise 5
3
8.9
.5
3
2.1
e
t
5
6.10
.6
32
22.2
uu
uu
ee
ee
t
t
3
2.11
1
2.7
2.3
2
2
s
s
s
2
1142
2
113012
28
422
4
2
.ln..
.
sincos
sin.
tt
11
