MA1200 Exercise for Chapter 7 Techniques of Differentiation
Brief Solutions
First Principle
17
1. a)
(3 x  4) 2
lim
b)
h 0
2( x  h )  1  2 x  1
2( x  h )  1  2 x  1

h
2( x  h )  1  2 x  1
2h
2


h  0 h[ 2( x  h)  1  2 x  1]
2 2x  1
1
 f ' ( x)
2x  1
 lim
Product/Quotient/Chain Rules
1
6
6
2. a) 28 x  12 x  1  2  3  4
x
x
x
x
c)
25  x 2
8x  x3
e)
(4  x 2 )1.5
3
1
23 3 3
b)
x
3
 12
d)
(3  2 x) 2
f) 3000(1  2 x  5 x 2 ) 2999 (2  10 x)
g) 2( x 2  4)(2 x)(2 x3  1)3  ( x 2  4) 2 3(2 x3  1) 2 (6 x 2 )
h) 6 sin 2 (2 x) cos(2 x)  15 x 2 sin( x3  1)
1
i)
x ln x ln(ln x)
k) cot x
1
m)  1 
x
x 3
o) e ( x cos x  3 x 2 cos x  x 3 sin x)
p)
j) ln x
l)  tan x
n)  e x sin(e x )
(3 x 2  5 x)e x (2 x  sin x  2  cos x)  e x (2 x  sin x)(6 x  5)
(3 x 2  5 x) 2
(n  1) x n  nx n 1  1
 x2  4x  6
r)
( x  1) 2
( x 2  3x)2
s) 2 cos(2 x) cos(3x)  3 sin(2 x) sin(3x)
 sin(ln(2 x)  1) 1
t) cos(ln[cos(ln(2 x)  1)]  1)
cos(ln(2 x)  1) x
q)
3. a) y ' x / y
c)
b) y '  
( x2  y2 ) x  y
x  ( x2  y2 ) y
1
2x  y
x  2y
dy
dy dt 9t 2 9
9
9
4. a)


 t  when t = 1. Tangent: y  5  ( x  3) , y  2.25 x  1.75 Normal:
dx dx
4
4t 4
4
dt
4
4
19
y 5
x
( x  3) , y 
9
9
3
17
dy
1 17
b)
when t = 4. Tangent: y  x  5 ,
2 t  
4
dx
t 4
4
135
Normal: y   x 
17
34
cot x 

5. a) ( x  1) cot x  csc 2 x ln( x  1) 
x  1 

x 
e x (4 x 3  4 x) 
b) ( x 4  2 x 2 )e e x ln( x 4  2 x 2 ) 
x 4  2 x 2 

c) (sin 5 x) x
2 2
[2 x ln(sin 5 x)  5( x 2  2) cot 5 x]  3
6. a) y ( n )  3(4n )e 4 x
n 

b) y ( n )  5(6n ) sin  6 x  7 

2 

c) y ( n )  2n 1 (1) n n!(2 x  1)  n 1
3!
 11!
11 n
x 3 n
6
 (11  n)! x
(3  n)!
1 n  3

11!
(n) 
11 n
when 4  n  11
d) y  
x
(11  n)!

12  n
0


Leibniz’s rule
e 2 x x 2 (3  2 x)
n 1
 2x
(n)
2
7. a) y  2e x(2 x  6 x  3)
when n  2
 n 3 2 x 3 3
2
2
n3
2 e [2 x  3(2 )nx  3(2)n(n  1) x  n(n  1)(n  2)]
b) y ( n )



2
(2 x  4) cos(2 x  1)  2( x  4 x  7) cos 2 x  1  2  when n  1



 n
n  2

( x  4 x  7 )
2 cos 2 x  1 
2 



(n  1) 


 n2 n 1 cos 2 x  1 
(2 x  4)

2 


(n  2) 


 n(n  1)2 n  2 cos 2 x  1 
when n  2


2


2
(1  x) 2 x(2  x) when n  1
c) y ( n )  
(1) n n!(1  x)  n 1 when n  2
Miscellaneous
8. If y  x sin x , prove that x 2 y '' 2 xy ' 2  x 2 y  0 .


Proof:
y  x sin x  y '  sin x  x cos x  y "  2 cos x  x sin x . Hence x 2 y '' 2 xy ' (2  x 2 ) y    0 .
9.
If u  ax 2  2bx  c , prove that
Proof:
1

du d  2
  (ax  2bx  c) 2  
dx dx 

d
2ax 2  3bx  c
( xu ) 
.
dx
u
ax  b
1
(ax 2  2bx  c) 2
d
ax  b
2ax 2  3bx  c
ax  b
( xu )  x
u  
. Then
.

dx
u
u
u
 1
if | x |  c

*10. Find the values of a and b (in terms of c) provided f ' (c) exists, where f ( x)   |x|
.
ax  b if | x |  c
Solution:
 1
if x  c
 x
 1
if | x |  c


 f ( x)  ax  b if  c  x  c
f ( x)   |x|
ax  b if | x |  c
 1
if x  c
 x

In order that f ' (c) exists, f (x) must be continuous at x  c .
1 1
That is, lim f ( x)  lim   lim f ( x)  ac  b  f (c) .
x c
x c x
c x c
1
1 1
cx
xc
 (ac  b)

f ( x )  f (c )
1
1
lim
 lim x
 lim x c  lim xc   lim xc   lim
 2 .
x c 
x

c
x

c
x

c
x

c
x

c
xc
xc
xc
xc
xc
xc
c
f ( x )  f (c )
ax  b  (ac  b)
a ( x  c)
lim
 lim
 lim
 a.
x c 
x c
x c
xc
xc
xc
1
f ( x )  f (c )
f ( x )  f (c )
That f ' (c) exists means lim
  2  lim
 a.
x c
x c
xc
xc
c
1

a 2

1
1
1
2
1

c
.
Therefore, a   2 . Putting in ac  b  , we have   b   b  . Hence 
c
c
c
c
c
 b 2

c
3
11. A function y of x is defined by the equation sin( x  y )  m sin y . Express y explicitly in terms of x.
dy
1  m cos x
Hence, or otherwise, show that

.
dx 1  2 m cos x  m 2
Proof:
sin( x  y )  m sin y  sin x cos y  cos x sin y  m sin y  tan y 
sin x
 sin x 
 y  tan 1 

m  cos x
 m  cos x 
Then

 sin x  
d  tan 1 

dy
1  m cos x
 m  cos x  
 
 
dx
dx
1  2m cos x  m 2
12. Let y   x  1
cot x
, find
dy
.
dx
Solution:
dy
cot x
cot x
cot x
   x  1 csc2 x ln  x  1 
 x  1
dx
x 1
13. Show that f ' ( x)  0 , where f ( x)  tan 1 x  tan 1
1
.
x
Solution:
f '( x) 
1
x2 1

 0.
1  x2 1  x2 x2
 x  t 2  2
where    t   .
14. Consider the parametric curve 
3
 y  t
dy d 2 y
,
and the equation of the tangent line to the curve at the point (3, 1).
Find
dx dx 2
Solution:
 dx d 2 y dy d 2 x 
dy
 dy 
 dy 



dy dt
d 2 y d  dy  d  dt  d  dt  dt  dt dt 2 dt dt 2  dt

 2    
 
 
2
dx dx
dx
dx  dx  dx  dx  dt  dx  dx 
 dx
 dx 




dt
 dt 
 dt 
 dt 


dy
3
.
The slope of the tangent line to the curve at the point  3,1 =
x 3 
dx
2
Therefore, the equation of the tangent line to the curve at the point  3,1 is: y  1 
3
 x  3 .
2
15. By Leibnitz’s theorem on repeated differentiation, find the nth derivatives of the function y = e x cos 2 x .
4
Solution:
By Leibnitz’s theorem on repeated differentiation, we have
n
 n  d k g d nk f
n
d0 f
d0g
dn
n!
, k  0,1, , n and 0!  1 .
,
where

f
,
g,  
fg




  k
n
nk
0
0
dx
dx
dx
dx
k  0  k  dx
 k  k ! n  k  !


n
n
n 
dn x

We have n e cos2 x    2k  e x cos 2 x 

2 
dx

k 0
k 
-End-
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