Question Bank
5. CONTINUITY AND DIFFERRENTIABILITY
ONE MARK QUESTIONS
1. Find the derivative of cos( x 2 ) with respect to x.
Sol:
let y  cos( x2 )
dy
d
  sin( x 2 ) ( x 2 )  2 x sin( x 2 )
dx
dx
2. Find the derivative of e x  e x  e x  e x  e x with respect to x.
2
Sol :
3
4
5
let y  e x  e x  e x  e x  e x
2
3
4
5
2
3
4
5
dy
 e x  2 xe x  3x 2e x  4 x3e x  5x 4e x
dx
3. Find the derivative of log(logx) with respect to x.
Sol :
y  log(log x)
dy
1 d
1

(log x) 
dx log x dx
x log x
4. Find the derivative of cos(sinx) with respect to x.
Sol :
y  cos(sin x)
dy
d
  sin(sin x) (sin x)   cos x sin(sin x)
dx
dx
5. Find the derivative of sec(tan x ) with respect to x
Sol :
y  sec(tan x )

 

sec tan x tan tan x sec2 ( x )
dy
d
 sec tan x tan tan x
(tan x ) 
dx
dx
2 x

 

6. Find the derivative of the function cos( x ) with respect to x.
Sol :
y  cos
 x
dy
  sin
dx
 x  dxd  x  
 sin
 x
2 x
1
7. If y  3e2 x  2e3 x find
Sol :
dy
dx
y  3e2 x  2e3 x
 
 
dy
d 2x
d 3x
3
e 2
e  6e2 x  6e3 x  6(e2 x  e3 x )
dx
dx
dx
8. Find the derivative of 5cosx – 3sinx with respect to x.
Sol :
y  5cos x  3sin x
dy
 5sin x  3cos x
dx
9. The function f ( x) 
Sol :
f ( x) 
1
is a quotient function. The function f(x) is not defined at x = 5 because
x 5
f (5) 
1
1
 is not defined. Therefore f(x) is continuous for all values of x except x = 5.
55 0
10. Find
Sol :
1
is not continuous at x = 5. Justify the statement.
x 5
dy
if x  y  
dx
x y 
d
d
 x  y    
dx
dx
d
d
( x)  ( y )  0
dx
dx
dy
1
dx
11. If y  tan(2 x  3) find
Sol :
dy
dx
y  tan(2 x  3)
dy
d
 sec2 (2 x  3)  2 x  3  2sec2 (2 x  3)
dx
dx
12. Find the derivative of f given by f ( x)  tan 1 x assuming it exists.
Sol :
y  tan 1 x

dy
1

dx 1  x 2
2
13. Prove that the function f ( x)  x n is continuous at x = n, where n is a positive integer.
f ( x)  x n , n  N .
Sol :
Here, f(x) is a polynomial function and D f  R
lim f ( x)  lim x n  nn  f (n).
x n
x n
Therefore f(x) is continuous at n  N .
14. Find the derivative of esin
Sol :
y  esin
1
1
x
with respect to x.
x
1
dy
esin x
sin 1 x d
1
e
sin x 
dx
dx
1  x2

15. Find
Sol :

dy
, if x  at 2 , y  2at .
dx
x  at 2 ,
y  2at
dx
 2at ,
dt
dy
 2a
dt
dy
dy dt
2a 1



dx dx 2at t
dt
TWO MARK QUESTIONS:
1. Examine whether the function f given by f ( x)  x 2 is continuous at x = 0.
Sol :
f ( x)  x 2 at x = 0;
f (0)  0 .
Then lim f ( x)  lim x 2  0
x 0
x 0
lim f ( x)  0  f (0) .
x 0
 f is continuous at x = 0.
2. Discuss the continuity of the function f defined by f ( x) 
Sol :
1
, x  0.
x
1 1

x c x
c
Fix any non zero real number c, we have lim f ( x)  lim
x c
3
1
Also, since for c  0 , f (c)  , we have lim f ( x)  f (c) and hence, f is continuous at every
x c
c
point in the domain of f. Thus f is continuous function.
3. Find the derivative of the function y 
Sol :
y
ex
with respect to x.
sin x
ex
sin x
dy

dx
sin x
 
d x
d
e  e x  sin x 
dx
dx
2
sin x
x
dy e x sin x  e x cos x e  sin x  cos x 


dx
sin 2 x
sin 2 x
4. Discuss the continuity of the function f given by f ( x)  x3  x 2  1
Sol :
Clearly f is defined at every real number c and its value at c is c3  c2  1 . We also know that


lim f ( x)  lim x3  x 2  1  c3  c 2  1
x c
x c
Thus lim f ( x)  f (c) , and hence f is continuous at every real number. This means f is a
x c
continuous function.
5. Verify Rolle’s theorem for the function y  x 2  2 , a = -2 and b = 2
Sol :
The function y  x 2  2 is continuous in [- 2 , 2] and differentiable in (-2, 2). Also
f (2)  f (2)  6 and hence the value of f ( x) at -2 and 2 coincide.  Rolle’s theorem states that
there is a point c  (2, 2) , where f | (c)  0 . Since f | ( x)  2 x , we get c = 0. Thus at c = 0, we
have f | (c)  0 and c  0  (2, 2) .
6. If f and g be two real functions continuous at real number c. Then show that f + g is continuous at
x = c.
Sol :
The continuity of f + g at x = c, clearly it is defined at x = c we have
lim  f  g  x   lim  f ( x)  g ( x)
x c
x c
 lim f ( x)  lim g ( x)  f (c)  g (c)   f  g  (c)
x c
x c
Hence f + g is continuous at x = c.
4
7. Find
dy
if, x  a cos , y  a sin  .
dx
Sol :
x  a cos ,
y  a sin 
dx
 a sin 
d
dy
 a cos 
d
dy
dy d
a cos 


  cot 
dx
dx
a sin 
d
8. Discuss the continuity of the function f given by f ( x) | x | at x = 0.
Sol :
 x, if x<0
By definition f ( x)  
 x, if x  0
Clearly the function is defined at x = 0 and f(0) = 0.
Let hand limit of f at x = 0 is lim f ( x)  lim(
 x)  0

x 0
x 0
Right hand limit of f at x = 0 is lim f ( x)  lim x  0 .
x 0
x 0
Thus the left hand limit, right hand limit and the value of the function coincide at x = 0.Hence, f is
continuous at x = 0.
9. Find the derivative of the function
Sol :
y
sin(ax  b)
with respect to x.
cos(cx  d )
sin(ax  b)
cos(cx  d )
dy

dx
cos(cx  d )
d
d
sin(ax  b)  sin(ax  b) cos(cx  d )
dx
dx
2
cos (cx  d )
dy a cos(cx  d ) cos(ax  b)  c sin(ax  b)sin(cx  d )

dx
cos 2 (cx  d )
10. Discuss the continuity of sine function.
Sol :
f ( x)  sin x is defined for every real number. Let c be a real number. Put x = c + h.
If x  c we know that h  0 . Therefore
lim f ( x)  limsin x
x c
x c
5
 limsin(c  h)  lim sin c.cosh  cos c.sinh 
h0
h0
 limsin c.cosh  limcos c.sinh  sin c  0  sin c  f (c)
h0
h0
Thus lim f ( x)  f (c)
x c
Therefore f is a continuous function.
11. Differentiate xsin x x  0 with respect to x.
Sol :
y  xsin x
Taking log on both sides
log y  log xsin x
log y  sin x log x
1 dy
d
d
 sin x (log x)  (sin x) log x
y dx
dx
dx
1 dy sin x

 cos x log x
y dx
x
dy
 sin x

 y
 cos x log x 
dx
 x

dy
 sin x

 xsin x 
 cos x log x 
dx
 x

12. Differentiate the function sin  tan 1 e x  with respect to x.
Sol :
y  sin  tan 1 e x 

 
dy
d
 cos tan 1 e x
tan 1 e x
dx
dx



1  x
x
dy  cos tan e e

dx
1  e2 x
13. If x  2at 2 , y  at 4 find
Sol :
dy
dx
x  2at 2 ,
y  at 4
dx
 4at ,
dt
dy
 4at 3
dt
6
dy
dy dt 4at 3


 t2
dx dx 4at
dt
14. If xy  e x  y , prove that
Sol :
dy y ( x  1)

dx x( y  1)
xy  e x  y
Differentiating both sides with respect to x.

d
d x y
( xy ) 
e
dx
dx

x
dy
 dy 
 y  e x  y 1  
dx
 dx 
x
dy
dy
 y  e x y  e x y
dx
dx
x
dy x  y dy
e
 e x y  y
dx
dx
e
 x  e  dy
dx
x y
x y
y
dy e x  y  y xy  y y ( x  1)



dx x  e x  y x  xy x( y  1)
15. If y  cos x cos 2 x cos3x find
Sol :
dy
dx
y  cos x cos 2 x cos3x
Taking log on both sides, we get
log y  log(cos x cos 2 x cos3x)
log y  log cos x  log cos 2 x  log cos3x
1 dy
sin x 2sin 2 x 3sin 3x



y dx
cos x cos 2 x cos 3x
dy
  cos x cos 2 x cos3x  tan x  2 tan 2 x  3tan 3x 
dx
16. If
x  y  a prove that
dy
y

dx
x
7
Sol :
x y a
Differentiate w.r.t x we get
1
2 x

1 dy
0
2 y dx
1 dy
1

2 y dx
2 x

dy
y

dx
x
17. Find the derivative of  sin x  cos x 
sin x cos x
Sol :
y   sin x  cos x 
with respect to x.
(sin x cos x )
Taking log on both sides
log y  log(sin x  cos x)(sin xcos x )
log y  (sin x  cos x)log(sin x  cos x)
Differentiate w.r.t x
1 dy
d
d
  sin x  cos x  log(sin x  cos x)  (sin x  cos x) log(sin x  cos x)
y dx
dx
dx
1 dy
cos x  sin x
 (sin x  cos x)
 (cos x  sin x) log(sin x  cos x)
y dx
sin x  cos x
dy
(sin x cos x )
  sin x  cos x 
 cos x  sin x 1  log(sin x  cos x)
dx

18. If y  sin 1 x
Sol :


x
y  sin 1 x

find
dy
dx
x
Taking logarithm on both sides

log y  log sin 1 x


x
log y  x log sin 1 x




1 dy
d
d
 x log sin 1 x  ( x) log sin 1 x
y dx
dx
dx

1 dy
x

 log sin 1 x

1
2
y dx sin x 1  x


8

dy
x

 y
 log sin 1 x

1
2
dx

 sin x 1  x

19. If y  sin  loge x  prove that
Sol :

 = sin x 
1

1 y2
dy

dx
x
y  sin  loge x 
cos  log e x 
dy
d
 cos  log e x   log e x  
dx
dx
x
1   sin(log e x 
1 y2
dy


dx
x
x
2
THREE MARK QUESTIONS:
 1  x2 
dy
1. If y  cos 1 
0 < x < 1 Find
2 
dx
 1 x 
Sol :
 1  x2 
y  cos 1 
2 
 1 x 
Put x  tan 
 1  tan 2  
1
y  cos 1 
  cos  cos 2 
2
 1  tan  
y  2  2 tan 1 x
Differentiate w.r. t x
dy
1

dx 1  x 2


2. If x3  x2 y  xy 2  y 3  81 . Find
Sol :
dy
dx
x3  x2 y  xy 2  y 3  81
Differentiate w.r. t x
3x 2  x 2
x
2
dy
dy
 dy 
 2 xy  x  2 y   y 2  3 y 2
0
dx
dx
 dx 
 2 xy  3 y 2
   3x
 dy
dx
2
 2 xy  y 2

9
x

x

 log sin 1 x
 1
2

 sin x 1  x





3x 2  2 xy  y 2
dy
 2
dx
x  2 xy  3 y 2
 x  3  x 2  4 
3. Diffentiate y 
Sol :
y

3x 2  4 x  5
w.r. t x.
 x  3  x 2  4 
3x 2  4 x  5
Taking logarithm on both sides, we have
log y 




1
log  x  3  log x 2  4  log 3x 2  4 x  5 

2
Differentiating on both sides w.r.t x, we get
1 dy 1  1
2x
6x  4 
 
 2
 2
y dx 2  x  3 x  4 3x  4 x  5 
dy y  1
2x
6x  4 
 
 2
 2
dx 2  x  3 x  4 3x  4 x  5 
dy 1

dx 2
 x  3  x 2  4  
1
2x
6x  4 
 2
 2

3x  4 x  5  x  3 x  4 3x  4 x  5 
2
4. Find
dy
 2x 
if y  sin 1 
2 
dx
 1 x 
Sol :
 2x 
y  sin 1 
2 
 1 x 
Put x  tan 
 2 tan 
y  sin 1 
2
 1  tan 

1
  sin  sin 2 

y  2  2 tan 1 x
Differentiating w.r.t x, we get
dy
2

dx 1  x 2
5. Differentiate the function  log x 
cos x
with respect to x.
10
y   log x 
Sol :
cos x
Taking logarithm on both sides
log y  log  log x 
cos x
log y  cos x log  log x 
Differentiating w.r. t. x on both sides, we get
1 dy
d
d
 cos x  log log x    cos x  log  log x 
y dx
dx
dx
1 dy cos x

 sin x log  log x 
y dx x log x
 cos x


dy
cos x  cos x
 y
 sin x log  log x    log x  
 sin x log  log x 
dx
 x log x

 x log x

6. Find
dy
if y x  x y
dx
Sol :
yx  xy
Taking logarithm on both sides
log y x  log x y
x log y  y log x
Differentiating with respect to x, on both sides, we get
x
d
d
d
d
(log y)  log y ( x)  y (log x)  log x ( y)
dx
dx
dx
dx
x dy
y
dy
 log y   log x
y dx
x
dx
 x  y log x  dy y  x log y

 
y
x

 dx
dy y  y  x log y 

dx x  x  y log x 
7. Differentiate sin 2 x with respect to ecos x
Sol :
let u  sin 2 x
and
v  ecos x
Differentiate w.r.t x
11
du
 2sin x cos x
dx
and
dv
  sin xecos x
dx
du
du dx 2sin x cos x
2 cos x


  cos x
cos x
dv dv  sin xe
e
dx
 sin x 
8. Differentiate tan 1 
 with respect to x.
 1  cos x 
Sol :
 sin x 
y  tan 1 

 1  cos x 
x
x
x


2sin cos 
sin 


2
2  tan 1
2  tan 1  tan x 
y  tan 1 




x
x
2

 2 cos 2

 cos 
2 
2


y
x
2
Differentiating w.r.t x on both sides
dy 1

dx 2
9. Verify mean value theorem if f ( x)  x3  5x 2  3x in the interval [a, b] where a = 1 and b = 3. Find
c  (1,3) for which f | (c)  0 .
Sol:
Given f ( x)  x3  5x 2  3x x [1,3] which is a polynomial function.
Since a polynomial function is continuous and derivable at all x  R
(1) f(x) is continuous on [1,3] (2) f(x) is derivable on (1, 3)
Therefore condition of mean value theorem satisfied on [1,3]. Hence,  at least one real c  (1,3)
3
2
f (3)  f (1) 3  5(3)  3(3)  1  5(1)  3(1)
f (c ) 

3 1
2
|
f | (c )  
20
 10
2
f | ( x)  3x 2  10 x  3 ;
f | (c)  3c2  10c  3  10
3c2  10c  7  0
3c2  7c  3c  7  0 ;
12
c(3c  7)  (3c  7)  0  c = 1  (1,3)
c
7
 (1,3) .
3
Hence the mean theorem satisfied for given function in the given interval.
10. If y  cos1 x find
Sol :
d2y
in terms of y alone.
dx 2
y  cos1 x
x  cos y
Differentiating w.r.t y, we get
dx
  sin y
dy
dy
  cos ecy
dx
Again differentiating w.r. t x , we get
d2y
dy
 cos ecy cot y
2
dx
dx
d2y
  cos ec 2 y cot y
2
dx
11. Find the derivative of  log x 
Sol :
y   log x 
log x
with respect to x.
log x
Taking logarithm on both sides
log y  log  log x 
log x
log y  log x log  log x 
Differentiating w.r. t x on both sides
1 dy
d
d
 log x  log  log x    log  log x   log x 
y dx
dx
dx
1 dy
log x log  log x 


y dx x log x
x
 1 log  log x  
log  log x  
dy
log x  1
 y 
   log x   

dx
x
x
x

x

13
12. Find
Sol :
 
dy
if y  cos x3 .sin 2 x5
dx
y  cos  x3  sin 2  x5 
Differentiating w.r.t x on both sides
    sin  x  dxd  cos x 
dy
d
 cos x3
sin 2 x5
dx
dx

2
5

3


dy
 cos x3 2sin x5 cos x5 5 x 4  sin x5  sin x3 3x 2
dx
dy
 10 x 4 sin x5 cos x5 cos x3  3x 2 sin x5 sin x3
dx
13. Verify Rolle’s theorem for the function f ( x)  x 2  2 x  8 , x [4, 2] .
Sol :
Given f ( x)  x 2  2 x  8 , x [4, 2] . Since a polynomial function is continuous and derivable on
R. (1) f(x) is continuous on [-4,2]
(2) f(x) is derivable on [-4,2].
Also f (4)  (4)2  2(4)  8  0 and f (2)  22  2  2  8  0  f (4)  f (2) .
This means that all the conditions of Rolle’s theorem are satisfied by f(x) in [-4,2].
Therefore there exist at least one real number c  (4, 2) such that f | (c)  0 .
f ( x)  x 2  2 x  8  f | ( x )  2 x  2
f | (c)  0  2c  2  0  c  1 (4, 2)
 Rolle’s theorem is verified with c = - 1
14. If x  asin
Sol :
1
x  asin
xa
and y  acos
t
1
t
1 1
sin t
2
1
t
then prove that
y  acos
ya
1
dy
y

dx
x
t
1
cos 1 t
2
Differentiating w.r.t “t” we get
1 1
sin t
dx
d 1

 a2
log a  sin 1 t 
dt
dx  2

1 1
sin t
2
dx a
log a

dt
2 1 t2
1
cos 1 t
dy
d 1

 a2
log a  cos 1 t 
dt
dx  2

1
cos1 t
2
dy
a
log a

dt
2 1 t2
14
1
1
dy a 2 cos t  log a
1
2
dy dt
a sin t
2
1

t

 1 1

1
sin t
dx dx
log
a
a cos t
a2

dt
2 1 t2
dy
y

dx
x
15. Find the derivative of x x  2sin x with respect to x.
Sol :
let y  x x  2sin x = u – v
Where u  x x and v  2sin x
Taking log on both sides
log u  log x x
and
log v  log 2sin x
log u  x log x
and
log v  sin x log 2
Differentiate with respect to x we get
1 du
d
d
 x  log x    x  log x
u dx
dx
dx
and
1 dv
 cos x log 2
v dx
du
x

 u   1.log x   x x 1  log x 
dx
x

and
dv
 v cos x log 2  2sin x cos x log 2
dx
y  u v
dy du dv


 x x 1  log x   2sin x cos x log 2
dx dx dx
dy

 tan
dx
2
16. If x  a(  sin  ) and y  a(1  cos ) prove that
Sol :
x  a(  sin  )
y  a(1  cos )
Differentiating w.r.t x on both sides
dx
 a(1  cos  )
d
dy
 a(0  sin  )  a sin 
d
dy
dy d
a sin 



dx dx a (1  cos  )
d
dy

dx
2a sin

2
cos
2a cos 2

2

2 
sin

2  tan 

2
cos
2
17. If a function f(x) is differentiable at x = c , prove that it is continuous at x = c.
15
Sol :
Since f is differentiable at c, we have lim
x c
But for x  c , we have f ( x)  f (c) 
f ( x )  f (c )
 f | (c )
xc
f ( x )  f (c )
. x  c 
xc
 f ( x )  f (c )

Therefore lim[ f ( x)  f (c)]  lim 
.  x  c 
x c
x c
xc


Or
 f ( x )  f (c ) 
lim f ( x)  lim f (c)  lim 
 x  c
 lim
x c
x c
x c
x c
xc

 f | (c) . 0 = 0
lim f ( x)  f (c) . Hence f is continuous at x = c.
x c
FIVE MARK QUESTIONS
1. If y  3cos  log x   4sin  log x  prove that x 2 y2  xy1  y  0 .
Sol :
y  3cos  log x   4sin  log x 
Differentiating w.r.t x on both sides
y1  
3sin  log x 
x

4cos  log x 
x
xy1  3sin  log x   4cos  log x 
Again differentiating on both sides we get
xy2  1 y1  
3cos  log x 
x

4sin  log x 
x
x2 y2  xy1   3cos  log x   4sin  log x 
x 2 y2  xy1   y
 x2 y2  xy1  y  0
2. If y  3e2 x  2e3 x prove that
Sol :
d2y
dy
5  6y  0
2
dx
dx
y  3e2 x  2e3 x

dy
 6e2 x  6e3 x  6 e2 x  e3 x
dx

d2y
 2  12e2 x  18e3 x  6 2e2 x  3e3 x
dx


16
Hence
d2y
dy
 5  6 y  6(2e3 x  3e3 x)  30 e2 x  e3 x  6 3e2 x  2e3 x
2
dx
dx

 

d2y
dy
 5  6 y  12e3 x  18e3 x  30e2 x  30e3 x  18e2 x  12e3 x  0
2
dx
dx

3. If y  tan 1 x
Sol :

2

y  tan 1 x


2


prove that x 2  1 y2  2 x x 2  1 y1  2 .

2
Differentiating w.r.t on both sides
y1 
2 tan 1 x
1  x2


 1  x 2 y1  2 tan 1 x
Again differentiating w.r.t x on both sides
1  x  y
2
2
 2 xy1 
2
1  x2
On cross multiplication, we get
1  x 
2
2


y2  2 x 1  x 2 y1  2
4. If y  Aemx  Benx , Show that
Sol :
d2y
dy
  m  n   mny  0 .
2
dx
dx
y  Aemx  Benx
Differentiating w.r.t x on both sides
dy
 Amemx  Bnenx Again differentiate w.r.t x on both sides
dx
d2y
 Am2emx  Bn2enx
2
dx
d2y
dy
  m  n   mny  m2 Aemx  n2 Benx   m  n  Amemx  Bnenx  mny
Hence
2
dx
dx


 m2 Aemx  n2 Benx  Am2emx  Bmnenx  Amnemx  n2 Benx  mny
  Bmnenx  Amnemx  mny  mn  Aemx  Benx   mny

 mny  mny  0
17
y  Aemx  Benx


5. If y  sin 1 x prove that 1  x 2
Sol :
2
0
 ddxy  x dy
dx
2
y  sin 1 x
Differentiate w.r.t x, we get
dy
1

dx
1  x2
1  x2
On cross multiplication
dy
1
dx
Again Differentiate w.r.t x, we get
1  x2
d2y
2 x dy

0
2
dx
2 1  x 2 dx
Taking Lcm and simplifying, we get

1  x2

d2y
dy
x
0
2
dx
dx


6. If y  cos1 x prove that 1  x 2 y2  xy1  0
Sol :
y  sin 1 x
Differentiate w.r.t x, we get
dy
1

dx
1  x2
On cross multiplication
1  x 2 y1  1
Again Differentiate w.r.t x, we get
1  x 2 y2 
2x
2 1  x2
y1  0


Taking Lcm and simplifying, we get 1  x 2 y2  xy1  0
7. If y  5cos x  3sin x , prove that
Sol :
d2y
y0
dx 2
y  5cos x  3sin x
Differentiating w.r.t x , on both sides
18
dy
 5sin x  3cos x
dx
Again differentiating w.r.t x we get
d2y
 5cos x  3sin x    5cos x  3sin x 
dx 2
d2y
 y
dx 2

d2y
y0
dx 2
d 2 y  dy 
8. If e  x  1  1 , prove that
 
dx 2  dx 
2
y
Sol :
e y  x  1  1
Differentiate w.r.t x on both sides
ey
 
d
d
 x  1   x  1 e y  0
dx
dx
e y   x  1 e y
dy
0
dx
dy
1

dx
x 1
Again differentiate w.r.t x , we get
d2y
1

2
2
dx
 x  1
d 2 y  dy 
 
dx 2  dx 
2
9. If y  500e7 x  600e7 x then show that
Sol :
d2y
 49 y
dx 2
y  500e7 x  600e7 x
Differentiate w.r.t x



dy
 500 7e7 x  600 7e7 x
dx

Again differentiate w.r.t x
d2y
 500 49e7 x  600 49e7 x
dx 2




d2y
 49 500e7 x  600e7 x
dx 2

d2y
 49 y
dx 2
19


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