ma2176a5
Derivative and Differentiation
1.
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 "  cos x  cos x  x sin x  2 cos x  x sin x .
So
x 2 y' '2 xy'(2  x 2 ) y  x 2 2 cos x  x sin x   2 xsin x  x cos x   2  x 2 x sin x
 x 3  sin x  sin x   x 2 2 cos x  2 cos x   x 2 sin x  sin x   0
2.
If u  ax 2  2bx  c , prove that
Proof:
du d

dx dx

 ax
2

 2bx  c 
ax  b
1
2



d
2ax 2  3bx  c
( xu) 
dx
u
1
1

 1
d  2
 (ax  2bx  c) 2   (ax 2  2bx  c) 2 (2ax  2b)
 2
dx 

ax  b
u
(ax 2  2bx  c)
Then
d
du
ax  b
ax 2  bx  u 2
( xu)  x
u  x
u 
dx
dx
u
u
2
2
2
ax  bx  ax  2bx  c 2ax  3bx  c


u
u
3.
Find from first principle the derivative of 2 x sin 2 x .
Solution:
d  2 x sin 2 x 
2( x  h )sin 2( x  h )  2 x sin 2 x
 lim
h 0
dx
h
 2 x  2h  sin 2 x cos 2h  cos 2 x sin 2h   2 x sin 2 x
(2 x  2h )sin(2 x  2h )  2 x sin 2 x
 lim
 lim
h 0
h

0
h
h
2 x sin 2 x cos 2h  2h sin 2 x cos 2h  2 x cos 2 x sin 2h  2h cos 2 x sin 2h  2 x sin 2 x
 lim
h 0
h
 2 x sin 2 x  cos 2h  1 2h sin 2 x cos 2h 2 x cos 2 x sin 2h 2h cos 2 x sin 2h 
 lim 




h 0
h
h
h
h


cos 2h  1
sin 2h
 2sin 2 x lim cos 2h  4 x cos 2 x lim
 2 cos 2 x lim sin 2h
h

0
h

0
h 0
h
2h
2sin 2 h
sin h
 2 x sin 2 x lim
 2sin 2 x  4 x cos 2 x  2 x sin 2 x lim  2sin h  lim
 2sin 2 x  4 x cos 2 x
h 0
h 0
h 0
h
h
 2sin 2 x  4 x cos 2 x
 2 x sin 2 x lim
h 0
4.
Differentiate the following functions with respect to x:
1
(a)
1  2 x  5x 
(b)
sin 3 (2 x)  5 cos( x 3  1)
(a)
2 3000
ln  ln  ln x  
Solution:
(a)
3000
2999
d
1  2 x  5x 2 
 3000 1  2 x  5 x 2   2  10 x 

dx
(b)
d
sin 3 2 x  5 cos  x 3  1  3  sin 2 2 x  (2)  5sin  x 3  1 3 x 2   6sin 2 2 x  15 x 2 sin  x 3  1
dx


(c)
d
 ln x 
d
1
dx
ln(ln
x
)


d
1
x
 ln x 

ln  ln(ln x )   dx
dx
ln  ln x 
ln  ln x  ln x ln  ln x  x ln x ln  ln x 
5.
Determinate the derivative g (x ) in terms of f   x 
(a)
g ( x )  f  x 2  1
(b)
g ( x)  f  f ( x) 
Solution:
(a)
g ( x )  f  x 2  1  g '( x )  f '  x 2  1  2 x   2 xf '  x 2  1
(b)
g ( x)  f  f ( x)   g '( x)  f '  f ( x)  f '( x)
6.
Suppose f is a differentiable function. Find
d
f (2 x)
(a)
dx
d
f (sin 2 x)
(b)
dx
d
f  x 2  1
(c)
dx
d
[ f ( x )] 2
(d)
dx
Solution:
(a)
d
f (2 x)  2 f ' (2 x)
dx
2
(b)
d
f (sin 2 x)  2(cos 2 x) f ' (sin 2 x)
dx
(c)
d
f  x 2  1  2 xf '  x 2  1
dx
(d)
d
 f ( x )2  2 f ( x ) f '( x )
dx
7.
Find the values of a and b (in terms of c) such that f ' (c ) exists, where
 1
if | x |  c

f ( x)   |x|

ax  b if | x |  c
Solution:
 1
if x  c
 x
 1
if | x |  c


f ( x)   |x|
 f ( x)  ax  b if  c  x  c
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 )
lim
 lim x
 lim x c  lim xc   lim xc
x c 
x

c
x c x  c
x c x  c
x c x  c
xc
xc
1
1
  lim
 2
x c xc
c
f ( x )  f (c )
ax  b  (ac  b)
lim
 lim
x c 
x

c
xc
xc
a ( x  c)
 lim
a
x c
xc
f ( x )  f (c )
1
f ( x )  f (c )
  2  lim
 a.
That f ' (c ) exists means lim
x c
x c
xc
xc
c
1
1
Therefore, a   2 .and put in ac  b  .
c
c
3
1
1 c 1
We have   b  c 2  b  c 2  
.
c
c
c
So b  c 2 .
3
1

 a   c 2
Finally, we have 
3
b  c  1

c
A function y of x is defined by the equation sin( x  y )  m sin y . Express y explicitly
dy
1  m cos x

in terms of x. Hence, or otherwise, show that
.
dx 1  2 m cos x  m 2
8.
Proof:
sin( x  y )  m sin y  sin x cos y  cos x sin y  m sin y
 sin x cos y  m  cos x sin y

sin y
sin x
sin x

 tan y 
cos y m  cos x
m  cos x
 sin x 
 y  tan 1 

 m  cos x 
Then

 sin x  
d  tan 1 
 
dy
 m  cos x  


dx
dx
=
  sin x  
d  
 
2
  m  cos x   (m  cos x ) cos x  sin x
(m  cos x) 2
dx


2
m  cos x 2  sin 2 x
 sin x 
1 

(m  cos x) 2
 m  cos x 
1  m cos x
1  2 m cos x  m 2
9.
Let f be a differentiable function. Prove that if f is odd, then f ' is even.
Proof:
f (  x  h)  f (  x )
f  ( x  h)   f ( x)
f ' ( x)  lim
 lim
h0
h0
h
h
 f  x  (  h)   f ( x ) 
 f ( x  h)   f ( x) 
  f  x  (  h)   f ( x ) 
 lim
 lim
 lim
h0
h0
h0
h
h
h
 f ' ( x)
10.
Let y   x  1
cot x
, find
dy
.
dx
Solution:
4
y   x  1

11.
cot x
dy
 1 
 ln y  cot x ln  x  1  dx   csc x ln  x  1  cot x 

y
 x 1
dy
cot x 
cot x
cot x
cot x

 y   csc x ln  x  1 
 x  1
    x  1 csc x ln  x  1 
dx
x 1 
x 1

Show that f ' ( x)  0 , where f ( x)  tan 1 x  tan 1
1
.
x
Solution:
1
d 
1
1
1
 x
f ( x)  tan 1 x  tan 1  f ' ( x) 

2
1 dx
x
1 x
1 2
x
2
1
x
1
1
1




0
2
2
2
2
1 x
1 x x
1 x
1 x2
12.
Differentiate tan 1 (ln x) with respect to x.
Solution:
d
1
 ln x 
d
1
1
x
 tan (ln x )   dx


2
2
2
dx
1   ln x 
1   ln x 
x 1   ln x  


 x  t 2  2
Consider the parametric curve 
where    t  
3
 y  t
dy d 2 y
,
Find
and the equation of the tangent line to the curve at the point  3,1 .
dx dx 2
Solution:
Method 1:
2
2
3

x  t  2
3
2 .

x

y

2

y

x

2



3

y  t
Then
3
1
1
1
dy 3
d2y  1  3
3


2
2
y   x  2 
  x  2 , 2     x  2 2   x  2 2
dx 2
dx
4
22
Method 2:
 dx d 2 y dy d 2 x 
dy
 dy 
 dy 


2
2
 d 
 dv 
dy dt
d y d  dy  d 
dt dt 2  dt

 2      dt    dt    dt dt
2
dx dx
dx
dx  dx  dx  dx  dt  dx  dx 
 dx
 dx 
 


dt
 dt 
 dt 
 dt 


13.
5
d 2x
 dx

2
t
2
 x  t  2  dt
dy 3t 2 3
 dt 2


.
So

 t



3
2
dx 2t 2
 y  t
 dy  3t 2
 d y  6t
 dt 2
 dt
 dx d 2 y dy d 2 x 
2



2
d y  dt dt 2 dt dt 2  dt  2t  6t   2  3t   dt 3 dt
 


2
 dx 2 dx
dx 2 
4t 2
 dx 
 dx 


 


dt




2
1
dt
dt 1


t  0 .
dx
dx 2t


1
1
dy 3
3
d 2 y 3 dt 3  1  3 
1
  3  x  2  2
 t   x  22 ,



As a result,
 
1
2
dx 2
2
dx
2 dx 2  2t  2  2 x  2 2  4
 
 
dy
3
The slope of the tangent line to the curve at the point  3,1 =
.
x 3 
dx
2
3
Therefore, the equation of the tangent line to the curve at the point  3,1 is: y  1   x  3
2
In addition, x  t 2  2  t   x  2  2 also x  t 2  2  1  2t
14.
Let y  v 2  v, v 
x
d2y
, find
.
x 1
dx 2
Solution:
dy dy dv

dx dv dx
d 2 y d  dy  d  dy dv  dy d  dv  dv d  dy 
   

 
 
dx 2 dx  dx  dx  dv dx  dv dx  dx  dx dx  dv 
2
dy d 2v dv d  dy  dv dy d 2v d 2 y  dv 


  
 
dv dx 2 dx dv  dv  dx dv dx 2 dv 2  dx 
 d 2 v  x  1  x
 dv  x  1  x
2
1




x


2
3
2

2
2
x  1
x  1
v 
 dx
 dx
x  1
x  1





x 1  

 y  v 2  v  dy
d2y

2
v

1
2
 dv
 dv 2
Therefore,
2
d 2 y dy d 2v d 2 y  dv 
2
1


2
   2v  1
3
4
2
2
2 
dx
dv dx
dv  dx 
 x  1
 x  1


2  2v  1
 x  1
3

2
 x  1
4
 2x

2
 1
4 x  2  x  1  2 6 x  4
2
x 1 
 



3
4
4
4
 x  1
 x  1
 x  1
 x  1
6
By Leibniz’s theorem on repeated differentiation, we have
n
dn
d n f n1
d i g d n i f
dng
d i g d n i f
, where
fg

g

C

f

C


 n i dxi dxni dxn 
n i
dx n
dx n i 1
dxi dx n i
i 0
15.
d0 f
d 0g

f
,
 g.
dx 0
dx 0
Find the nth derivatives of the following functions:
e x cos 2 x
(a)
x 3e2 x
(b)
1
(c)
 x  1 2 x  1
Solution:
(a)
di f
f  x   e x  i  e x , i  0,1,2,
dx
d 2i g
d 2i 1g
i 2i
i 1
g  x   cos 2 x  2i   1 2 cos 2 x,
  1 22i 1 sin 2 x, i  0,1,2,
2 i 1
dx
dx
So
n i
x
n
d i  cos 2 x  d  e 
dn x
 e cos 2 x    n Ci dxi
dx n
dx n i
i 0
If n  2m where m is a nonnegative integer, then
n i
x
n
d i  cos 2 x  d  e 
dn x
e
cos
2
x

C

  n i dxi
dx n
dx n i
i 0
n 2i
n  2 i 1
e x  m 1
ex 
d 2i  cos 2 x  d
d 2i 1  cos 2 x  d


  n C2 i
  n C2i 1
dx 2i
dx n 2i
dx 2i
dx n 2i
i 0
i 0
m
m
m 1
  n C2i  1 22i  cos 2 x  e x   n C2i 1  1 22i 1  sin 2 x  e x
i
i 0
i 1
i 0
If n  2m  1 , where m is a nonnegative integer, then
n i
x
n
d i  cos 2 x  d  e 
dn x
e
cos
2
x

C

  n i dxi
dx n
dx n i
i 0
n 2i
n  2 i 1
ex  m
ex 
d 2i  cos 2 x  d
d 2i 1  cos 2 x  d


  n C2 i
  n C2i 1
dx 2i
dx n 2i
dx 2i
dx n 2i
i 0
i 0
m
m
m
  n C2i  1 22i  cos 2 x  e x   n C2i 1  1 22i 1  sin 2 x  e x
i
i 0
i 1
i 0
(b)
3i
d i f  3 Px
i
f  x  x  i  
dx
 0
3
i  0,1,2,3
i  4,5,
, where 3 Pi  i ! 3 Ci
7
dig
 2i e2 x , i  0,1,2,
i
dx
g  x   e2 x 
So,
n
d i  e2 x  d n i  x 3  n
d n 3 2x
If n  3 , n  x e    n Ci
  n Ci 2i e2 x 3 Pn i x 3n i
i
n i
dx
dx
dx
i 0
i 0
If n  3 ,
n
n
d i  e 2 x  d n i  x 3 
d i  e 2 x  d n i  x 3 
d n 3 2x
 x e    n Ci dx i dx ni   n Ci dx i dx ni
dx n
i 0
i  n 3
n


i  n 3
i 2x
3 n i
n Ci 2 e
3 Pn i x

j  i   n  3
3
 n Cn3 j 2 j n3 e2 x 3 P3 j x j
j 0

n Cn  3 j  n C3 j
3

j 0
n
C3 j 2 j n 3 e 2 x 3 P3 j x j
(c)
1
d n  1   1 n !
f  x 
 n

x 1
dx  x  1   x  1n 1
n
n
1
d n  1   1 2 n !
g  x 
 n

, n  0,1, 2,

2x  1
dx  2 x  1   2 x  1n 1
So
 1  n i  1 
di 
n
d 

n

d 
1
2x  1 

 x 1

C


 n i dx i
dx n   x  1 2 x  1  i 0
dx n i
n
 1 2i i !  1  n  i !  n n !
 1 2i i !  n  i !
  n Ci
C

n i
i 1
n i 1
i 1
n i 1
i 0
i 0 i !  n  i  !
 2 x  1  x  1
 2 x  1  x  1
n
n
1 2i n !


i 1
n i 1
i 0  2 x  1
 x  1
n
i
n i
n
-End-
8