Methods of Differentiation

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METHODS OF DIFFERENTIATION
The process of calculating derivative is called differentiation.
1. DERIVATIVE OF f(x) FROM THE FIRST PRINCIPLE :
y f(x  x)  f(x) dy
Obtaining the derivative using the definition Lim  Lim  f '(x)  is
x 0 x x 0 x dx
called calculating derivative using first principle or ab initio or delta method.
dy dy
Note : can also be represented as y1 or y' or Dy or ƒ '(x). represents instantaneous rate of
dx dx
change of y w.r.t. x.

Illustration 1 : Differentiate each of following functions by first principle :

(i) f(x) = tanx (ii) f(x) = esinx
tan(x  h)  tan x tan(x  h  x) 1  tan x tan(x  h) 
Solution : (i) f'(x) = lim = lim
h 0 h h 0 h
tanh
= lim . (1 + tan2x) = sec2x. Ans.
h 0 h
esin(x  h)  esin x esin(x  h)sin x  1  sin(x  h)  sin x 
(ii) f'(x) = lim = lim esin x  
h 0 h h 0 sin(x  h)  sin x  h 
sin(x  h)  sin x
= esin x lim = esinxcosx Ans.
h 0 h
Do yourself -1 :
(i) Differentiate each of following functions by first principle:
1
(a) f(x) = nx (b) f(x) =
x

2. DERIVATIVE OF STANDARD FUNCTIONS :

f(x) f'(x) f(x) f'(x)
(i) xn nxn–1 (ii) ex ex
(iii) ax axna, a > 0 (iv) nx 1/x
(v) logax (1/x) logae, a > 0, a 1 (vi) sinx cosx
(vii) cosx – sinx (viii) tanx sec2x
(ix) secx secx tanx (x) cosecx – cosecx . cotx
(xi) cotx – cosec2x (xii) constant 0
1 1
(xiii) sin–1 x , 1  x  1 (xiv) cos–1 x , 1  x  1
1  x2 1  x2
1 1
(xv) tan–1 x , xR (xvi) sec–1 x , | x | 1
1  x2 | x | x2  1
1 1
(xvii) cosec–1 x , | x | 1 (xviii) cot–1 x , xR
| x | x2  1 1  x2

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3. FUNDAMENTAL THEOREMS :
If f and g are derivable functions of x, then,
d df dg d df
(a) (f  g)   (b) (cf)  c , where c is any constant
dx dx dx dx dx
d dg df
(c) (fg)  f g known as “PRODUCT RULE”
dx dx dx
 df   dg 
g   f  
d f  dx   dx  where g  0 known as “QUOTIENT RULE”
(d)   
dx  g  g2
dy dy du
(e) If y = f(u) & u = g (x) then  . known as “CHAIN RULE”
dx du dx
dy du
Note : In general if y = f(u) then  f '(u). .
dx dx

dy
Illustration 2 : If y = ex tan x + xlogex, find .
dx
Solution : y = ex.tan x + x · logex
On differentiating we get,
dy 1
= ex · tan x + ex · sec2x + 1 · log x + x ·
dx x
dy
Hence, = ex(tanx + sec2 x) + (logx + 1) Ans.
dx
log x dy
Illustration 3 : If y = + ex sin2x + log5x, find .
x dx
Solution : On differentiating we get,
1
·x  log x .1
dy d  log x  d x d x
    (e sin 2x)  (log 5 x) 
dx dx  x  dx dx x2
1
+ exsin2x + 2ex.cos2x +
x log e 5
dy  1  log x  x 1
Hence,   + e (sin2x + 2cos2x) + Ans.
dx  x 2
 x log e 5
dy
Illustration 4 : If y = loge (tan 1 1  x2 ) , find .
dx
Solution : y = loge (tan 1 1  x2 )
On differentiating we get,
1 1 1
= . . .2x
1
tan 1  x 1  ( 1  x ) 2 1  x2
2 2 2

x x
=  Ans.
 
 tan 1 1  x 2  1   1  x 2  1  x 2  tan 1 1  x 2   2  x 2  1  x 2
2

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Do yourself -2 :
dy
(i) Find if -
dx
(a) y = (x + 1) (x + 2) (x + 3) (b) y = e5x tan(x2 + 2)

4. LOGARITHMIC DIFFERENTIATION :
To find the derivative of a function :
(a) which is the product or quotient of a number of functions or
(b) of the form [f(x)]g (x) where f & g are both derivable functions.
It is convenient to take the logarithm of the function first & then differentiate.

dy
Illustration 5 : If y = (sin x)n x, find
dx
Solution :  n y = n x. n (sin x)
On differentiating we get,
1 dy 1 cos x dy  n(sin x) 
 n (sinx) + n x.  = (sinx)n x   cot x n x  Ans.
y dx x sin x dx  x 
 1  y  x 2   dy
Illustration 6 : If x = exp  tan  2   , then equals -
  x  dx
(A) x [1 + tan (log x) + sec2 x] (B) 2x [1 + tan (log x)] + sec2 x
(C) 2x [1 + tan (log x)] + sec x (D) 2x + x[1 + tan(logx)]2
Solution : Taking log on both sides, we get
 y  x2 
log x = tan–1  2   tan (log x) = (y – x2) / x2   y = x2 + x2 tan (log x)
 x 
On differentiating, we get
dy
 = 2x + 2x tan (log x) + x sec2 (log x)  2x [1 + tan (log x)] + x sec2 (log x)
dx
= 2x + x[1 + tan(logx)]2 Ans. (D)
x (1  2x)
1/2 2/3
dy
Illustration 7 : If y = find
(2  3x) (3  4x)
3/4 4/5
dx
1 2 3 4
Solution :  n y = n x + n (1 – 2x) – n (2 – 3x) – n (3 – 4x)
2 3 4 5
On differentiating we get,
1 dy 1 4 9 16
     
y dx 2x 3(1  2x) 4(2  3x) 5(3  4x)
dy  1 4 9 16 
     y        Ans.
dx  2x 3(1  2x) 4(2  3x) 5(3  4x) 

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Do yourself -3 :
dy dy 2 3 4
(i) Find if y = xx (ii) Find if y  e x .e x .e x .e x
dx dx

5. PARAMETRIC DIFFERENTIATION :
dy dy / d ƒ '()
If y  f() & x  g() where  is a parameter, then  
dx dx / d g'()

dy 
Illustration 8 : If y = a cos t and x = a(t – sint) find the value of at t =
dx 2
dy a sin t dy
Solution :    1 Ans.
dx a(1  cos t) dx t  
2

Illustration 9 : Prove that the function represented parametrically by the equations.

1 t 3 2
x  3 ;y  2 
t 2t t
dy
satisfies the relationship : x(y')3 = 1 + y' (where y' = )
dx
1 t 1 1
Solution : Here x =  3 2
t3 t t
Differentiating w.r. to t
dx 3 2
 4  3
dt t t
3 2
y 
2t 2 t
Differentiating w.r. to t
dy 3 2
 3  2
dt t t
dy dy / dt
  t  y'
dx dx / dt
1 t 1 y'
Since x = 3
 x or x(y')3 = 1 + y' Ans.
t (y ')3
6. DERIVATIVE OF A FUNCTION W.R.T. ANOTHER FUNCTION :
dy dy / dx f '(x)
Let y= f (x) ; z = g (x) then  
dz dz / dx g'(x)

Illustration 10 : Differentiate loge (tan x) with respect to sin–1(ex).

d
(log e tan x)
d(log e tan x) dx cot x.sec 2 x e  x 1  e 2x
Solution :    Ans.
d(sin 1 (e x )) d
sin 1 (e x ) e x .1 / 1  e 2x sin x cos x
dx

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Do yourself-4 :
dy 
(i) Find at t  if y = cos4t & x = sin4t .
dx 4
(ii) Find the slope of the tangent at a point P(t) on the curve x = at2 , y = 2at.
(iii) Differentiate xnx with respect to nx.

7. DIFFERENTIATION OF IMPLICIT FUNCTIONS : (x, y)  0

(a) To find dy /dx of implicit functions, we differentiate each term w.r.t. x regarding y as a
function of x & then collect terms with dy/dx together on one side.

dy 
(b) Also   x , where = partial derivative of (x, y) w.r.t. x taking y as a constant
dx  x
y

and = partial derivative of (x, y) w.r.t. y taking x as a constant.
y
(c) In the case of implicit functions, generally, both x & y are present in answers of dy/dx.

dy
Illustration 11 : If xy + yx = 2, then find .
dx
Solution : Let u = xy and v = yx
du dv
u+v=2    0
dx dx
Now u = xy and v = yx
 n u = y nx and n v = x n y

1 du y dy 1 dv x dy
  + nx and = n y +
u dx x dx v dx y dx
du y dy  dv  x dy 
 = xy   nx  and  yx  n y 
dx x dx  dx  y dx 

 x y
 y ny  x y . 
y dy   x dy  dy
    xy   n x  + yx  ny  = 0   x Ans.
x dx   y dx  dx  y x x
 x nx  y . y 
 
Aliter :
(x, y)  xy  yx  2  0

dy  / x yx y 1  y x ny
 
dx  / y x y nx  xy x 1

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sin x dy 1  y  cos x  y sin x

Illustration 12 : If y = , prove that  .
cos x dx 1  2y  cos x  sin x
1
sin x
1
1  cos x.....
sin x (1  y) sin x
Solution : Given function is y = =
cos x 1  y  cos x
1
1 y
or y + y2 + y cos x = (1 + y) sin x ......(i)
Differentiate both sides with respect to x,
dy dy dy dy
 2y  cos x  y sin x = (1 + y) cosx + sin x
dx dx dx dx
dy
(1 + 2y + cosx – sinx) = (1 + y) cosx + ysinx
dx
dy 1  y  cos x  y sin x
or  Ans.
dx 1  2y  cos x  sin x
Aliter :
From (i) (x,y) = (1 + y)sinx – y – y2 – ycosx = 0
dy  / x (1  y) cos x  y sin x (1  y) cos x  y sin x
  
dx  / y sin x  1  2y  cos x 1  2y  cos x  sin x

Do yourself -5 :
dy
(i) Find , if x + y = sin(x – y)
dx
(ii) If x2 + xey + y = 0, find y', also find the value of y' at point (0, 0).

8. DIFFERENTIATION BY TRIGONOMETRIC TRANSFORMATION :

Some Standard Substitutions :
Expression Substitution

a2  x2 x = asin or acos

a2  x2 x = tan or acot

x2  a2 x = asec or acosec

ax ax
or x = acos or acos2
ax ax

2ax  x 2 x = a(1 – cos)

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 2x 
Illustration 13 : If f(x) = sin–1   then find
 1  x2 

1
(i) f'(2) (ii) f'   (iii) f'(1)
2

 
Solution : x = tan, where    y = sin–1(sin2)
2 2

 
   2 2
 2  
    2 tan 1 x x 1
    1
y = 2   2   f(x) =  2 tan x 1  x  1
 2 2 (  2 tan 1 x)
 x  1
 
(  2)   2   2


 2
 1  x 2 x 1

 2
 f ' (x) =  1  x  1
 1 x
2

 2
 1  x 2 x  1

2 1 8
(i) f ' (2) =  (ii) f'    (iii) f ' (1+) = –1 and f ' (1–) = +1 f ' (1) does not exist Ans.
5 2 5

d  2  1 1  x   
Illustration 14 : sin  cot   =
dx 
  1  x  

1 1
(A)  (B) 0 (C) (D) –1
2 2

 1 x   
Solution : Let y = sin2  cot 1  . Put x = cos 2    0, 
 1  x   2


 1  cos 2 
 y = sin2 cot–1  = sin2 cot–1 (cot )
 1  cos 2 
 

1  cos 2 1  x 1 x
 y = sin2  =   
2 2 2 2
dy 1
  . Ans (A)
dx 2

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1  x2  1 1  1  x2
Illustration 15 : Obtain differential coefficient of tan–1 with respect to cos–1
x 2 1  x2
Solution : Assume u = tan–1, v = cos–1
  
The function needs simplification before differentiation Let x = tan     , 
 2 2
 sec   1  –1  1  cos     
 u = tan–1   = tan   = tan
–1
 tan  =
 tan    sin    2 2

1  sec  1  cos    
v = cos–1 = cos–1 = cos–1  cos  = u = v
2 sec  2  2 2
du
 = 1. Ans.
dv
Do yourself : 6
(i) If y = cos–1(4x3 – 3x), then find :
 3  3
(a) ƒ '   , (b) ƒ ' (0), (c) ƒ '  .
 2   2 
   

9. DERIVATIVE OF A FUNCTION AND ITS INVERSE FUNCTION :

If g is inverse of f, then
(a) g{f(x)} = x (b) f{g(x)} = x
g'{f(x)}f'(x)=1 f '{g(x)}g'(x) = 1

1
Illustration 16 : If g is inverse of f and f'(x) = , then g'(x) equals :-
1  xn
(A) 1 + xn (B) 1 + [f(x)]n (C) 1 + [g(x)]n (D) none of these
Solution : Since g is the inverse of f. Therefore
f(g(x)) = x for all x
d
   f(g(x))  1 for all x
dx
1
 f'(g(x)) g'(x) = 1  g'(x) = = 1 + (g(x))n Ans. (C)
f '(g(x))

Do yourself -7 :
(i) If g is inverse of ƒ and ƒ (x) = 2x + sinx; then g’(x) equals:
3 1 1
(A)   (B) 2 + sin–1x (C) 2 + cos g(x) (D)
x 2
1  x2 2  cos(g(x))

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10. HIGHER ORDER DERIVATIVES :
Let a function y = ƒ (x) be defined on an interval (a, b). If ƒ (x) is differentiable function, then its
derivative ƒ '(x) [or (dy/dx) or y'] is called the first derivative of y w.r.t. x. If ƒ '(x) is again
differentiable function on (a, b), then its derivative ƒ "(x) [or d2y/dx2 or y"] is called second
derivative of y w.r.t. x. Similarly, the 3rd order derivative of y w.r.to x, if it exists, is defined by
d3y d  d 2 y 
   and denoted by ƒ '''(x) or y''' and so on.
dx 3 dx  dx 2 
dy dy / d d 2 y d  dy  dx
Note : If x = f() and y = g() where '' is a parameter then  & 
dx dx / d dx 2 d  dx  d
d n y d  d n 1y  dx
In general    .
dx n d  dx n 1  d

11. ANALYSIS AND GRAPHS OF SOME INVERSE TRIGONOMETRIC FUNCTIONS :

2 tan 1 x | x | 1
 2x  
(a) y  f(x)  sin 1  2 
    2 tan 1 x x  1
 1  x   1
 (  2 tan x) x  1
Important points : y
   /2
(i) Domain is x  R & range is   , 
 2 2 I
(ii) f is continuous for all x but not differentiable
D
at x = 1,–1
2 x
 for | x |  1 –1 0 1
1  x
2

dy  D 1
(iii)   non existent for | x |  1
dx 
 2 –/2
 for | x |  1
1 x 2

(iv) Increasing in ( –1 , 1) & Decreasing in ( , 1)  (1, )

1  1  x2  2 tan 1 x if x0
(b) Consider y  f(x)  cos  2 
 
1 x 
1
 2 tan x if x0
Important points :
(i) Domain is x  R & range is [0, ) f(x)
(ii) Continuous for all x but not differentiable
at x = 0

2
 for x0 /2
1  x
2

dy 
(iii)   non existent for x0 x
dx  –1 0 1
 2
 for x0
1  x2
(iv) Increasing in (0, ) & Decreasing in (,0)

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2 tan 1 x | x | 1
12x 
(c) y  f(x)  tan     2 tan 1 x x  1
1  x2  (  2 tan 1 x)
 x 1

Important points :
  
(i) Domain is R – {1, – 1} & range is   , 
 2 2
(ii) It is neither continuous nor differentiable f(x)
at x = 1, –1
/2
2
 | x | 1
dy 1  x 2
(iii)  x
dx 
 non existent | x |  1 –1 0 1

(iv) Increasing  x in its domain

/2
(v) It is bounded for all x

 1 1
 (  3sin x) if 1  x  
2

1 1
(d) y  f(x)  sin (3x  4x )  3sin 1 x
1 3
if  x
 2 2

   3sin 1 x 1
if  x 1
 2
Important points : y

(i) Domain is x  [ 1, 1] & range is

   D
 2 , 2  I
 
(ii) Continuous everywhere in its domain x
–1 0 1
1 1
(iii) Not derivable at x   ,
2 2 I
D
 3 1 1
 if x  ( , )
dy  1  x 2 2 2
(iv) 
dx  3 1 1
 if x  (1,  )  ( ,1)
 1 x
2 2 2

 1 1  1 1 
(v) Increasing in   ,  and Decreasing in  1,  2    2 ,1
 2 2    

12 E
 Methods of Differentiation

 1 1
3cos x  2  if 1  x  
2

1 1
(e) y  f(x)  cos1 (4x  3x)  2   3cos 1 x
3
if  x
 2 2

3cos1 x 1
if  x 1
 2
Important points :
(i) Domain is x  [ 1, 1] & range is [0, ]
(ii) Continuous everywhere in its domain
y


D I D

/2

x
–1 O 1

1 1
(iii) Not derivable at x   ,
2 2
 3  1 1
 if x    , 
dy  1  x 2  2 2
(iv) 
dx  3  1 1 
 if x   1,     ,1 
 1 x
2
 2 2 

 1 1
(v) Increasing in   ,  &
 2 2
 1 1 
Decreasing in  1,     ,1 
 2 2 
GENERAL NOTE :
Concavity is decided by the sign of 2nd derivative as :
d2y d2y
 0  Concave upwards ;  0  Concave downwards
dx 2 dx 2

E 13
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JEE-Mathematics

Illustration 17 : Find the interval for which f(x) = x3 + x + 1 is

(i) concave upwards (ii) concave downwards.
3
Solution : f(x) = x + x + 1
f'(x) = 3x2 + 1
f"(x) = 6x
(i) f"(x) = 6x > 0  Concave upwards
 x  [0, )
(ii) f"(x) = 6x < 0   Concave downwards
 x  (–, 0] Ans.
d2y
Illustration 18 : If x = a (t + sin t) and y = a(1 – cos t), find .
dx 2
Solution : Here x = a (t + sin t) and y = a (1–cos t)
Differentiating both sides w.r.t. t, we get :
dx dy
= a(1 + cos t) and = a (sin t)
dt dt
t t
2 sin .cos

dy

a sin t
 2 2  tan  t 
 
dx a 1  cos t  2 cos2
t 2
2
Again differentiating both sides, we get,
t
2 sec 2  
d y  t  1 dt 1 1 1 2
= sec2      sec 2  t / 2    
dx 2
 2  2 dx 2 
a 1  cos t  2a  2 t 
2  cos 
 2
d2y 1 t
Hence,   sec 4   Ans.
dx 2
4a 2
Illustration 19 : y = f(x) and x = g(y) are inverse functions of each other then express g'(y) and g''(y)
in terms of derivative of f(x).
dy dx
Solution :  f '(x) and  g '(y)
dx dy
1
 g'(y)  ...........(i)
f '(x)
Again differentiating w.r.t. to y
d  1  d  1  dx f ''(x)  1 
g ''(y)       .  .
dy  f '(x)  dx  f '(x)  dy (f '(x)) 2  f '(x) 
f ''(x)
 g''(y)   ...........(ii)
(f '(x))3
d2y
2
d x dx 2
Which can also be remembered as = – 3
Ans.
dy 2  dy 
 dx 
 

14 E
 Methods of Differentiation

Do yourself : 8
2
(i) If y = xe x then find y''.
(ii) Find y" at x = /4, if y = x tan x.
(iii) Prove that the function y = ex sin x satisfies the relationship y'' – 2y' + 2y = 0.

12. DIFFERENTIATION OF DETERMINANTS :

f(x) g(x) h(x)
If F(x)  l(x) m(x) n(x) , where f, g, h. l, m, n, u, v, w are differentiable functions of x
u(x) v(x) w(x)
f '(x) g'(x) h '(x) f(x) g(x) h(x) f(x) g(x) h(x)
then F '(x)  l(x) m(x) n(x)  l '(x) m '(x) n '(x)  l(x) m(x) n(x)
u(x) v(x) w(x) u(x) v(x) w(x) u '(x) v '(x) w '(x)

Note : Sometimes it is batter to expand the determinant first & then differentiate.

x x2 x3
Illustration 20 : If f(x) = 1 2x 3x 2 , find f'(x).
0 2 6x
x x2 x3
Solution : Here, f(x) = 1 2x 3x 2
0 2 6x
On differentiating, we get,
d d 2 d  3
(x) (x ) x x x2 x3
dx dx dx x x2 x3
d   d   d  2
  f'(x) = 1 2x 3x 2  1 2x 3x  1 2x 3x 2
dx dx dx
0 2 6x d   d   d  
0 2 6x 0 2 6x
dx dx dx
1 2x 3x 2 x x2 x3 x x2 x3
or f'(x) = 1 2x 3x 2  0 2 6x  1 2x 3x 2
0 2 6x 0 2 6x 0 0 6
As we know if any two rows or columns are equal, then value of determinant is zero.
x x2 x3
= 0 + 0 + 1 2x 3x 2  f'(x) = 6 (2x2 – x2)
0 0 6
Therefore, f'(x) = 6x2 Ans.

Do yourself : 9
x 2
2x x2 x3
e x
(i) If ƒ(x)  , then find ƒ '(1). (ii) If ƒ(x)  x 2  2x 1 3x  1 , then find ƒ ' (1).
nx sin x
2x 1  3x 2 5x

E 15
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JEE-Mathematics

13. ˆ
L ' HOPITAL ' S RULE :

0 
(a) This rule is applicable for the indeterminate forms of the type , . If the function f(x)
0 
and g(x) are differentiable in certain neighbourhood of the point 'a', except, may be, at the
point 'a' itself and g'(x)  0, and if
lim f(x)  lim g(x)  0 or lim f(x)  lim g(x)   ,
x a x a x a x a

f(x) f '(x)
then lim  lim
x a g(x) x a g '(x)
f '(x)
provided the limit lim exists (L' Hôpital's rule). The point 'a' may be either finite or
x a g '(x)
improper (+  or –).
0 
(b) Indeterminate forms of the type 0.  or –  are reduced to forms of the type or by
0 
algebraic transformations.
(c) Indeterminate forms of the type 1, 0 or 00 are reduced to forms of the type 0 ×  by
taking logarithms or by the transformation [f(x)](x) = e(x).nf(x).

sin x
Illustration 21 : Evaluate lim x
x 0

log e x
sin x lim
Solution : lim x  lim e sin x loge x
e x0 cosecx

x 0 x 0

1/x
lim
= ex 0  cosecxcot x (applying L'Hôpital's rule)
2
sin 2 x  sin x   x 
lim  lim      2  0 
=e x 0 x cos x
e x 0  x   cos x 
 e 1  e0  1 Ans.
Illustration 22 : Solve lim logsin x sin 2x.
x0

Solution : Here lim logsin x sin 2x

x0

log sin 2x   
= lim  form 
x 0 log sin x   
1
 2 cos 2x
= lim sin 2x {applying L'Hôpital's rule}
x 0 1
 cos x
sin x
  2x  
    cos 2x cos 2x
= lim 
sin 2x 
 lim 1 Ans.
x 0  x  x 0 cos x
  cos x
 sin x 

16 E
 Methods of Differentiation

1/ n
 en 
Illustration 23 : Evaluate lim   .
n    

1/ n
 en 
Solution : Here, A = lim   (0 form)
n    

1  en  n log e  log    
 log A = lim log    lim  form 
n  n  n  n  
log e  0
= lim {applying L'Hôpital's rule}
n  1
1/ n
1  en 
logA = 1  A = e or lim   = e Ans.
n    

Do yourself : 10
tan x  x ex  x  1
(i) Using L' Hopital
ˆ 's rule find (a) lim (b) lim
x 0 x3 x 0 x2
sin x  tan x 1 n(1  x)
(ii) Using L' Hopital
ˆ 's rule verify that : (a) lim 3
= (b) lim 1
x 0 x 2 x  0 x

Miscellaneous Illustrations :
Illustration 24 : Find second order derivative of y = sinx with respect to z = ex.
dy dy / dx cos x
Solution :   x 
dz dz / dx e
d 2 y d  cos x  dx e x sin x  cos xe x 1
     .  . x
dz 2 dx  e x  dz e x
2
 
e


d2y

 sin x  cos x  Ans.
2
dz e 2x
 x  y  ƒ(x)  ƒ(y)
Illustration 25 : Let a function ƒ satisfies ƒ    x, y  R and ƒ '(0) = a, ƒ (0) = b,
 2  2
then find ƒ (x) hence find ƒ ''(x).
 x  y  ƒ(x)  ƒ(y)
Solution : ƒ 
 2  2
Diff. w.r.t. 'x'
xy 1 1  dy 
ƒ '  .  ƒ '(x)  x & y are independent to each other,   0
 2  2 2  dx 
xy
ƒ '   ƒ '(x)
 2 

E 17
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JEE-Mathematics

x
Let x = 0 & y = x ƒ '    ƒ '(0)  a
2

 ƒ '(x) = a

On integrating, we get ƒ (x) = ax + b ( ƒ (0) = b)

 ƒ ''(x) = 0
1 x 1 x 1 x 1
Illustration 26: Prove that 2
sec2  4 sec2 2  6 sec2 3  ..... = cosec2 x  2
2 2 2 2 2 2 x
x x x
Solution : Let cos .cos 2 .cos 3 .......
2 2 2
x x x x
= lim cos .cos 2 .cos 3 .......cos n
n  2 2 2 2
sin x sin x x x x sin x
 lim    cos .cot 2 cos 3 ........ 
n  x x 2 2 2 x
2 n sin
2n

 x  x   x
 n  cos   n  cos 2   n  cos 3   ....  n sin x  nx
 2  2   2 

Diff. w.r.t. x

1 x 1 x  1
  tan  2 tan 2  ......   cot x 
2 2 2 2  x

Diff. w.r.t. x again

1 x 1 x 1 x 1
2
sec 2  4 sec 2 2  6 sec 2 3  ......  cosec 2 x  2
2 2 2 2 2 2 x
Hence proved

18 E
 Methods of Differentiation

Elementary Problems on Differentiation
Differentiate the given functions with respect to x :
x 1
1. y 2. y  tan 3 x  tan x  x 3. y  xsec2 x  tan x
sin x  cos x 3
x x 1
4. y  a cos 5. y  tan 6. y  1  2 tan x
3 2
x
7. y  cos3 4x 8. y  tan 9. y  sin 1  x 2
2
 1
10. y  cot 3 1  x2 11. y  (1  sin 2 x)4 12. y  1  tan  x  
 x
1 x
13. y  cos2 14. y  sin2 (cos3x) 15. y  x arcsin x
1 x
arcsin x 1
16. y 17. y  (arcsin x)2 18. y
arccos x arcsin x
x
19. y  xsin x arctan x 20. y  (arccos x  arcsin x)n 21. y  arctan x
1  x2
14
22. y  arctan x2 23. y arcsin x 2  2x
2
 b  a cos x 
24. y  arccos   ; (a, b > 0, sinx > 0) 25. y  x 2 log3 x
 a  b cos x 
nx
26. y nx 27. y  xsin x nx 28. y
xn
1  nx
29. y 30. y  n(x2  4x) 31. y  n tan x
1  nx
x3
32. y  log2 [log3 (log5 x)] 33. y  n arctan 1  x 2 34. y  3 n sin
4
x3  2x cos x
35. y  x.10x 36. y 37. y
ex ex
x
1  10 x
38. y2 nx
39. y 40. y  3sin x
1  10 x
2 2 2 x
41. y  ae  b x 42. y  Ae  k x sin(x  ) 43. y  xx
44. y  (sin x)cosx 45. y  ( nx)x 46. y  x nx
(x  1)3 4 x  2 1
47. y 48. y  x sin x 1  e x 49. y  xx
5
(x  3)2
3 x(x 2  1)
50. y  2x x
51. y  (x2  1)sin x 52. y
(x 2  1)2
 1 
53. Prove that the function y = n   satisfies the relationship xy'+1= ey (where dash denotes derivative)
1 x 
arcsin x
54. Prove that the function y  satisfies the relationship (1 – x2)y' – xy = 1(where dash
1 x 2

denotes derivative).
E 19
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JEE-Mathematics

EXERCISE # (O-1)
Single Correct Choice Type
1 1 1 dy
1. If y    
    
   
, then is equal to-
1 x x 1 x x 1 x x dx

(A) 0 (B) 1 (C) ( +  + )x++–1 (D)

DF0001
Ans. A
1 1 1
Sol. y   
x x  x x  x x 
1    1    1   
x x x x x x

x x x
 y  
x   x  x  x   x  x  x   x  x 

x   x  x 
 y  y=1
x   x  x 

dy
 0
dx

1 x dy
2. If y = , then equals-
1 x dx

y y y y
(A) (B) (C) (D)
1  x2 x 1
2
1  x2 y 1
2

DF0005
Ans. B
1
 1  x 2
Sol. y   
 1 x 
1  1 x 
ny  n 
2  1 x 
1
ny = (n(1–x) – n(1 + x))
2

y ' 1  1 1 
    
y 2  1 x 1 x 

y ' 1  2 
   
y 2  1  x2 
y
 y' 
x 1
2

20 E
 Methods of Differentiation

dy
3. If x3 – y3 + 3xy2 – 3x2y + 1 = 0, then at (0, 1) equals-
dx
(A) 1 (B) –1 (C) 2 (D) 0
DF0012
Ans. A
Sol. x3 – y3 + 3xy2 – 3x2y + 1 = 0
(x – y)5 + 1 = 0
Differentiate wrt. x :
3(x – y)2 (1 – y') + 0 = 0
x = 0, y = 1
 3(1 – y') = 0
 y' = 1
  
4. If y2 + loge (cos2 x) = y, x    ,  , then :
 2 2
(A) |y"(0)| = 2 (B) |y'(0)| + |y"(0)| = 3
(C) |y'(0)| + |y"(0)| = 1 (D) y"(0) = 0 DF0173
Ans. A
Sol. y2 + loge(cos2x) = y …(1)
Put x = 0 y2 – y = 0
y = 0 or y = 1
Differentiate equation (1)
2yy' – 2tanx = y' …..(2)
Put x = 0 y'(x) = 0 for both y = 0 and y = 1
Differentiate equation (2)
2yy" + 2(y')2 – 2sec2x = y"
Put x = 0 y"(0) = –2 for y = 0
y"(0) = 2 for y = 1
|y"(0)| = 2
dy
5. If (cosx)y = (siny)x, then equals-
dx
log sin y  y tan x log sin y  y tan x log sin y  y tan x log sin y  y tan x
(A) (B) (C) (D)
log cos x  x cot y log cos x  x cot y log cos x  x cot y log cos y  y cot x
DF0008
Ans. B
Sol. (cosx)y = (sin y)x

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JEE-Mathematics

Taking log on both sides :

yn(cos x) = xn(sin y)

y( sin x) x(cos y)

 y ' n(cos x)  y ' n(sin y)
cos sin y
n sin y  y tan x
 y' 
n cos x  x cot y
dy
6. If 2x + 2y = 2x+y, then is equal to-
dx

2x  2y 2x  2y  2y 1  2x  y  2x
(A) (B) (C) 2x–y  x 
(D)
2x  2y 1  2x  y  1 2  2y
DF0009
Ans. C
Sol. 2x + 2y = 2x+y
2xn2 + (2yn2)y1 = 2x+y(n2)(1 + y1)

 2x + 2yy1 = 2x+y(n2)(1 + y1)

 y1(2y – 2x+y) = 2x+y – 2x

2x y  2x
 y1  
2y  2x y
2 x (2 y  1)
  y1  
2 y (1  2 x )

(2 y  1)
  y  2
1 x y

(1  2x )
7. If x2 + y2 = 1, then-
(A) yy" – 2(y')2 + 1 = 0 (B) yy" + (y')2 + 1 = 0
(C) yy" + (y')2 – 1 = 0 (D) yy" + 2(y')2 + 1 = 0
DF0018
Ans. B
Sol. x2 + y2 = 1
 2x + 2yy' = 0
 x + yy' = 0
 1 + (y')2 + yy" = 0
dy
For x >1, if (2x)2y = 4e2x–2y, then 1  loge 2x 
2
8. is equal to :
dx

22 E
 Methods of Differentiation

x log e 2x  log e 2 x log e 2x  log e 2
(A) loge2x (B) (C) xloge2x (D)
x x
DF0104
Ans. D
Sol. (2x)2y = 4e2x–2y
2yn2x = n4 + 2x – 2y

x  n2
y
1  n2x
1
(1  n 2x)(x  n 2)
y'  x
(1  n2x)2

 x n 2x  n 2 
y '(1  n 2x)   
 x

 dy d 2 y 
9. If ey + xy = e, the ordered pair  , 2  at x = 0 is equal to :
 dx dx 
 1 1 1 1  1 1  1 1
(A)   , 2  (B)  , 2  (C)  ,  2  (D)   ,  2 
 e e  e e  e e   e e 
DF0106
Ans. A
Sol. ey + xy = e
differentiate w.r.t. x
dy dy
ey x y 0 …..(1)
dx dx

dy dy 1
(x  e y )  y, 
dx dx (0,1) e

again differentiate w.r.t. x in (1)

d 2 y dy y dy d 2 y dy dy
ey .  .e .  x.   0
dx 2 dx dx dx 2 dx dx
2
d 2 y  dy  y dy
(x  e y ) 2
   .e  2 0
dx  dx  dx

d2 y 1  1
e 2
 2 e  2   0
dx e  e

d2 y 1
 2
 2
dx e

E 23
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JEE-Mathematics

10. Let y = y(x) be a function of x satisfying y 1  x 2  k  x 1  y 2 where k is a constant and

1 1 dy 1
y     . Then at x = , is equal to :
2 4 dx 2
5 5 2 5
(A) (B)  (C) (D)  DF0154
2 2 5 4
Ans. B
Sol. Put x = sin, y = sin

y 1  x2  k  x 1  y2

 sin.cos + cos.sin = k

 sin(.cos) = k

   +  = sin–1k

  sin–1k + sin–1y = sin–1k

1 1 dy
    0
1  x2 1  x2 dx

1 1
  at x = ,y=
2 4

dy  5

dx 2

11. Suppose the function f (x) – f (2x) has the derivative 5 at x = 1 and derivative 7 at x = 2. The
derivative of the function f (x) – f (4x) at x = 1, has the value equal to
(A) 19 (B) 9 (C) 17 (D) 14
DF0023
Ans. A
Sol. (f(x) – f(2x))' = f'(x) – 2f'(2x)|x=1 = 5
f'(1) – 2f'(2) = 5 ……(1)
at x = 2  f'(2) – 2f'(4) = 7 ……(2)
(f(x) – f(4x))' = f'(x) – 4f'(4)
at x = 1  f'(1) – 4f'(1)
equation (1) + 2 × equation (2)
f'(1) – 4f'(4) = 19

24 E
 Methods of Differentiation

12. Suppose that f (0) = 0 and f '(0) = 2, and let g (x) = f  x  f  f(x)   . The value of g ' (0) is equal

to
(A) 0 (B) 1 (C) 6 (D) 8
DF0028
Ans. C
Sol. g(x) = f(–x + f(f(x)))
g'(x) = f'(–x + f(f(x)) (–1 + f'(f(x))f'(x))
g'(0) = f'(f(f(0)) (–1 + f'(f(0))f'(0))
= f'(0)(–1 + f'(0)f'(0))
= 2(–1 + 22)
g'(0) = 6
 1  x2  1   2x 1  x 2  1
13. The derivative of tan–1   with respect to tan–1   at x = is :
 x   1  2x 
2
2

3 3 2 3 2 3
(A) (B) (C) (D) DF0174
12 10 5 3
Ans. B
   2 
Sol. Let y1 = tan–1  1  x  1  and y2 = tan–1  2x 1  x 
2

 x   1  2x 2 

 sec   1  1  
Let x = tan y1 = tan–1    tan  tan 
 tan    2

y1 
2
1
y1  tan 1 (x)
2
 2 sin  cos  
Let x = sin y2 = tan 1  1
  tan (tan 2)
 cos 2  
y2 = 2 = 2tan–1(x)
1 1
dy1 dy1 / dx 1  x 2 . 2 1  x2
  
dy 2 dy 2 / dx 2. 1 4(1  x 2 )
1  x2

1 dy1 3
x 
2 dy 2 10
dy
14. If y = tan–1(cotx) + cot–1(tanx), then is equal to-
dx

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JEE-Mathematics

(A) 1 (B) 0 (C) –1 (D) –2

DF0015
Ans. D
Sol. y = tan–1(cotx) + cot–1(tanx)
 
y= –cot–1(cotx) + – tan–1(tanx)
2 2
 y = p – 1((x + )) + (x + )
(where a, b depends on the interval of x)
y' = –2
 1  sin x  1  sin x    dy 5
15. If y(x) = cot 1   , x   ,   , then at x  is :
 1  sin x  1  sin x  2  dx 6

1 1
(A)  (B) –1 (C) (D) 0 DF0175
2 2
Ans. A
 1  sin x  1  sin x 1  sin x  1  sin x 
Sol. y(x) = cot–1   
 1  sin x  1  sin x 1  sin x  1  sin x 

 
y(x) = cot–1  2  2 1  sin x 
2

 2 sin x 

 2  2 | cos x |   
y  (x)  cot 1   x  , 
 sin x  2 

 x
y(x)  tan 1  cot 
 2
 x
y(x)  
2 2
1
y '(x)  
2
16. Let ƒ and g be differentiable functions on R such that fog is the identity function. If for some a,
b  R, g'(a) = 5 and g(a) = b, then ƒ'(b) is equal to :
2 1
(A) (B) 1 (C) (D) 5 DF0172
5 5
Ans. C
Sol. f(g(x)) = x
f '(g(x)).g'(x) = 1
Put x = x

26 E
 Methods of Differentiation

1
f '(g(a)) =
g '(a)
1
f ' (b) =
5
1
17. If g is the inverse of f and f'(x) = then g'(x) is equal to-
1  x3
1 1 1
(A) 1 + [g(x)]3 (B) (C) (D)
2(1  x 2 ) 2(1  x 2 ) 1   g  x  
3

DF0017
Ans. A
Sol. g is inverse o f :
g(f(x)) = x and f(g(x)) = x
 g'(f(x)).f '(x) = 1

1
 g'(f(x)) =
f '(x)

1
 g'(f(x)) = 1  1  x3
(1  x 3 )

Replace x  g(x)

g'(f(g(x))) = 1 + (g(x))3

18. Let ƒ : R  R be a differentiable function satisfying ƒ'(3) + ƒ'(2) = 0.

1
 1  ƒ  3  x   ƒ  3  x
Then lim   is equal to
x 0  1  ƒ  2  x   ƒ  2  
 
(A) e2 (B) e (C) e–1 (D) 1
DF0171
Ans. D
Sol. f '(3) + f '(2) = 0
1/x
 1  f(3  x)  f(3) 
lim  , from 1
x 0 1  f(2  x)  f(2) 
 
f (3 x)f (3) f (2)  f (2 x) 
lim   
 e x0 x x 


ef '(3)+f '(2) = e0 = 1

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JEE-Mathematics

 1 1  x2 
19. The value of Lim  1
  is equal to
x 0
 xsin x x2 

5 5 6 6
(A)  (B) (C) (D) 
6 6 5 5
DF0087
Ans. B
x  (1  x 2 )sin 1 x
Sol. lim
x 0 x 2 sin 1 x
x  sin 1 x  x 2 sin 1 x
 lim
x 0 sin 1 x
x2 x
x
x  sin 1 x  x 2 sin 1 x
 lim
x 0 sin 1 x
x2 x
x
x  sin 1 x  x 2 sin 1 x
 lim
x 0 x3
Applying
1 x2
1   2x sin 1 x
 lim 1 x 2
1 x 2
 
x 0 3x 2
 2
1  1  x  2x sin 1 x
  lim 1  x2 
x 0 3x 2

1  1  x 2  2x sin 1 x
 lim 
x 0 3x 2
1 1 2x
x    (2x)  2 sin 1 x 
2 1 x 2
1  x2
  lim
x 0 6x
3 3x 1 2
     (2x) 
(1  x )
2 3/2
 2 
lim 1  x 1  x2 
2

x 0 6
5
 
6

20. The value of Lim

x n  x2  1  x  is equal to
3
x 0 x

28 E
 Methods of Differentiation

1 1 1
(A) (B)  (C) (D) 6
6 6 12
DF0088
Ans. A
1  2x 
1   1
Sol. lim
 x 1  x  2 x 1 
2 2

x 0 3x 2
1
1
 lim x2  1
x 0 3x 2
1
1  2x
2  (x  1)3/2
2
 lim
x 0 6x
1
 lim
x 0 6(x  1)3/2
2

1

6

EXERCISE # (O-2)
Single Correct Choice Type
u(x) u '(x)  u(x)  '
1. Let u(x) and v(x) are differentiable functions such that  7 . If  p and   q,
v(x) v '(x)  v(x) 
pq
then has the value equal to -
pq
(A) 1 (B) 0 (C) 7 (D) –7
DF0021
Ans. A
u(x)
Sol. 7
v(x)
u'(x)
p
v'(x)

 u '(x)  '
 v '(x)   q
 
u(x) = 7v(x)
u'(x) = 7v'(x)
u '(x)
7q
v '(x)

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JEE-Mathematics

d  u(x)  d
 (7)
dx  v(x)  dx

q=0
pq 70
 1
pq 70
2. Let f (x) be a polynomial function of second degree. If f (1) = f (–1) and a, b, c are in A.P., then f
'(a), f '(b) and f '(c) are in
(A) G.P. (B) H.P. (C) A.G.P. (D) A.P.
DF0022
Ans. D
Sol. Let f(x) = Ax2 + Bx + C
f(1) = f(–1)
A+B+C=A–B+C
B=0
f(x) = Ax2 + C
f'(x) = 2ax
a, b, c are in A.P.
2Aa, 2Ab, 2Ac are in A.P.
f'(a), f'(b), f'(c), are in A.P.
d2x
3. If y = x + ex then is :
dy 2
ex ex 1
(A) ex (B)  (C)  (D)
(1  e x )3 (1  e x )2 (1  e x )3
DF0030
Ans. B
Sol. y = x + ex
dy
 1  ex
dx
dy 1

dx 1  e x
d2 x 1 dx
 .e x .
dy 2
(1  e )
x 2
dy

d2 x e x

dy 2 (1  e x )3

30 E
 Methods of Differentiation

3
d 2 x  dy  d 2 y
4. If + = K then the value of K is equal to
dy 2  dx  dx 2
(A) 1 (B) –1 (C) 2 (D) 0
DF0033
Ans. D
Sol. we know that
d2 x d2y
 
dy 2 dx 2
3
 dy 
 
 dx 
3
d 2 x  dy  d 2 y
   0
dy 2  dx  dx 2
k=0
 1 
5. Let f (x) = x + sin x. Suppose g denotes the inverse function of f. The value of g'    has
4 2
the value equal to
2 1
(A) 2 1 (B) (C) 2  2 (D) 2 1
2
DF0036
Ans. C
  1
Sol. f(x) = x + sinx f  
4 4 2
1 1
g'(f(x)) = 
f '(x) 1  cos x

Put x =
4
 1  1
 g'   
4 2  1  cos   
 
4

1 2 2 1
  

1  cos   2 1 2 1
4

2 2

f(4)  f(x 2 )
6. If f is differentiable in (0, 6) & f '(4) = 5, then Limit =
x 2 2x

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JEE-Mathematics

(A) 5 (B) 5/4 (C) 10 (D) 20

DF0029
Ans. D
f(4)  f(x 2 )  0 
Sol. lim  form 
x 2 2x 0 
Apply L'hospital Rule
1  f '(x 2 ).2x
lim
x 2 0 1
4f'(4)
= 20
(x  h)ƒ(x)  2hƒ(h)
7. Let ƒ(x) be differentiable at x = h, then Lim is equal to -
xh xh
(A) ƒ(h) + 2hƒ'(h) (B) 2ƒ(h) + hƒ'(h)
(C) hƒ(h) + 2ƒ'(h) (D) hƒ(h) – 2ƒ'(h)
DF0034
Ans. A
(x  h)f(x)  2hf(h) 0 
Sol. lim  from 
x h xh 0 
Applying L'hospital Rule
(x  h)f '(x)  f '(x)  0
lim
x h 1 0
2hf'(h) + f(h)
Multiple Correct Choice Type
4
8. If ƒ(x) = (2x – 3)5 + x  cos x and g is the inverse function of ƒ, then
3
7 3 7
(A) g'(2) = (B) g'(2) = (C) g''(2) = (D) g''(2) = 0
3 7 3
DF0037
Ans. B,D
4
Sol. If f(x) = (2x – 3)5 + x + cos x and g is the inverse function of f, then find right answer :
3
4
f(x) = (2x – 3x)5 + x + cosx
3
 3  1
f    2 , g '(f(x)) 
 2  f '(x)

32 E
 Methods of Differentiation

1
g '(f(x)) 
4
5(2x  3)4 2   sin x
3
3
Put x =
2
  3   1
g' f   
  2   5(2. 3  3)4 .2  4  sin  3 
 
2 3  2 
1 3
g'(2) = 
4 7
1
3
1
g'(f(x)) =
f '(x)
1
g"(f(x))f'(x) = – .f "(x)
f '(x)2
1
g"(f(x)) = .f "(x)
f '(x)3

1  3  
g"(2 )  20.2(2x  3) .2  cos  2  
3
3

4   
  1
3 
g"(2) = 0
 B&D
9. If ƒ(x) = x.|x|, then its derivative is :
(A) 2x (B) –2x (C) 2|x| (D) 2xsgnx
DF0038
Ans. C,D
Sol. If f(x) = x, |x| then its derivative is
 x2 , x  0
f(x) =  2
x , x  0
 2x , x  0
f '(x)  
2x , x  0
f'(0+) = 0, f'(0–) = 0
 f'(0) = 0
 f'(x) = 2|x|
= 2x sgn(x)

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JEE-Mathematics

 C&D
x   x
10. Lim is equal to -
x  x x  
e 
(A) log e   (B) log e  
 e
(C) tan(cot–1(n) – cot–1(1)) (D) tan(tan–1(1) – tan–1(n))

DF0042
Ans. A,C,D
x   x
Sol. Lim x
x  x  

x 1   n 
Lim x
x  x (1  n x)

n  
e
1 
.   n  1  n  e
      loge  
x (1  n x)
x
1 n  n(e) 

 nge(e/)

(C) tan[cot–1(n) – cot–1(1)]

  1  
tan  tan 1  1
  tan (1) 
  n 

 1 
 1 n   1 
tan  tan
1 
 1 
 n
1 n  e
 log e  
1 n  

(D) tan[tan–1(1) – tan–1(n)]

1 n 
1 n 
e
log e  

 ACD

34 E
 Methods of Differentiation

EXERCISE # (O-3)
Numerical Grid Type
1. Let f, g and h are differentiable functions. If f(0) = 1 ; g(0) = 2 ; h(0) = 3 and the derivatives of
their pair wise products at x = 0 are (f g)' (0) = 6 ; (g h)' (0) = 4 and (h f)' (0) = 5 then the value of
(fgh)'(0) is
DF0046
Ans. 16
(fg)'(0)h(0)  (fh)'(0)g(0)  (gh)'(0)f '(0)
Sol. (fgh)'(0) =
2

6  3  5  2  4 1
  16
2
2. A twice differentiable function f(x) is defined for all real numbers and satisfies the following
conditions
f(0) = 2; f '(0) = –5 and f "(0) = 3.
The function g(x) is defined by g(x) = eax + f (x)  x  R, where 'a' is any constant, if a1, a2, ..., an
n
are the values of 'a' such that g'(0) + g"(0) = 0 then the value of a
i 1
2
i is

DF0139
Ans. 5
Sol. g(x) = eax + g(x)

g'(x) = aeax + f'(x)

g'(0) = ae0 + f'(0) = a + (–5)

=a–5

g"(x) = a2eax + f"(x)

g"(0) = a2 + f"(0) = a2 + 3

g'(0) + g"(0) = 0

a – 5 + a2 + 3 = 0

a2 + a – 2 = 0

a = –2, 1
3. Let f(x) be a polynomial function such that f(2x) = f '(x) f "(x), then the value of f(3) is :-
DF0140
Ans. 12
Sol. Let degree of f(x) be n

 Degree of f(2x) = n

E 35
 
JEE-Mathematics

Degree of f'(x) = n – 1

Degree of f"(x) = n – 2

Since f(2x) = f'(x)f"(x)

 Degree of LHS n = Degree of RHS

= n – 1 + n – 2 = 2n – 3

 n = 2n – 3 n=3

Let f(x) = ax3 + bx2 + cx + d (a  0)

f(2x) = 8ax3 + 4bx2 + 2cx + d

f'(x) = 3ax2 + 2bx + c

f"(x) = 6ax + 2b

8ax3 + 4bx2 + 2cx + d = (3ax2 + 2bx + c)(6ax + 2b)

Equating the coefficients of x3, x2, x and constant, we get

4
8a = 18a2  a = [ a  0]
9

4b = 6ab + 12ab = 18ab

 a = 2/9 or b = 0

b=0 [ a = 4/9]

2c = 4b2 + 6ac  2c = 6ac

1
 a= or c = 0
3

c=0[ a = 4/9]

d = 2bc = 0

4 3
 f(x)  x
9
1 1
 d2y dy
4. If 2x = y 5  y 5
then (x2 – 1) 2
x  ky , then the value of 'k' is :-
dx dx
DF0076
Ans. 25

36 E
 Methods of Differentiation

1 1

Sol. y 5  y 5
 2x ……(1)

 1 1 1 1  1 1 
  y5  y 5  y '  2
5 5 

1 1  15 1  15 
 y y  y   2
5  5 

 y '  y 5  y  5   10y 
1 1

 

  1
 y '2 y 5  y  5   100 y 
1 2 2

  y '2  5 
 y y

5
1
  4  100 y 
1 2 2

 4y'2[x2 – 1] = 100y2

y'2(x2 – 1) = 25y2

y'2.2x + (x2 – 1)2y'.y" = 50 y.y'

 (x2 – 1)y" + xy' = 25y

 k = 25
5. The function f: R  R satisfies f(x2) · f"(x) = f'(x) · f'(x2) for all real x. Given that f(1) = 1 and
f"'(1) = 8, then the value of f'(1) + f"(1) is :-
DF0079
Ans. 6
Sol. f(x2).f"(x) = f'(x).'(x2) x R …….(1)

 f(x2).f"'(x) + f"(x).f'(x2).2x

 f'(x)f"(x2).2x + f'(x2).f"(x)

Put x = 1

 f(1).f"'(1) + f"(1).f'(1).2 = f'(1).f"(1).2 + f'(1).2 + f'(1).f"(1)

 1.8 = '(1).f"(1)

 f'(1).f"(1) = 8 …….(2)

Put x = 1 in equation (1)

E 37
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JEE-Mathematics

 f(1).f"(1) = f '(1).f '(1)

 f"(1) = [f '(1)]2 …….(3)

From equation (2) and (3)

[f '(1)]3 = 8  f '(1) = 2

 f"(1) = 4

 f '(1) + f"(1) = 6
6. Let f : R  R is a function such that f(x) = x3 + x2 f'(1) + x f"(2) + f"'(3) for all x  R, then the
value of f(2) – f(1) + f(0) is
DF0141
Ans. 0
Sol. f(x) = x3 + x2f '(1) + xf "(2) + f "'(3)  x R ……(1)

 f '(x) = 3x3 + xf ' (1) + f "(2)  x R ……(2)

 f "(x) = 6x + 2f ' (1)  x R ……(3)

 f '"(x) = 6  x R ……(4)

  f '"(3) = 6

Put x = 2 in (3)

 f "(2) = 12 + 2f ' (1)

Put x = 1 in (2)

 f "(1) = 3 + 2f ' (1) + f "(2)

 f '(1) = 3 + 2f ' (1) + 12 + 2f '(1)

  3f '(1) = –15

  f ' (1) = –5

  f "(2) = 12 + 2f '(1) = 12 + 2(–5)

  f "(2) = 2

  f(x) = x3 – 5x2 + 2x + 6

f(2) = 8 – 20 + 4 + 6 = –2

f(1) = 1 – 5 + 2 + 6 = 4

38 E
 Methods of Differentiation

f(0) = 6

 f(2) = f(1) – f(0)

 g(x), x0

7. Let g(x) be a polynomial, of degree one & f(x) be defined by f(x) =  1  x 1/x
, x0
 2  x 

  9 
If f(x) be a continuous function satisfying f'(1) = f (–1), then the value of 6  f  3   n    is
  4 
DF0142
Ans. 2
Sol. g(x) = ax + b

 ax  b x0

f(x) =  1  x 1/x
  x0
 2  x 

f(0+) = f(0) = f(0)


1
  =b=b
2

b=0
1/x
 1 x 
In vicinity of '1' say f(x) =  
2x

 1 x 
y 
2x

1
In y = (ln(1 + x) – ln(2 + x))
x

1 1 ln(1  x) 1 1 ln(2  x)
.  . 
1 (2  x) x x2
y '  1 x x 2 x
2

y x x2

At 1

 11  2
y=  
 2 1 3

2  1  1  ln 2 
1 1
  ln 3 
y'   2 1 1
3 3 1 
 1 1 
E 39
 
JEE-Mathematics

21  2 
y '    ln   
36  3 

f(–1) = –a + b (b = 0)

= –a

2  1 3
y'(1) = f(–1)  a   ln 
3 6 2

 2  1 3
 3  6  ln 2  x, x  0

 1/x
  1 x  , x0
  2  x 

8. Let f : R  R be defined as f(x) = x3 + 3x2 + 6x – 5 + 4e2x and g(x) = f –1 (x), then the value of
28(g'(–1)) is
DF0144
Ans. 2
Sol. f(x) = x3 + 3x2 + 6x – 5 + 4e2x

1
g'(f(x)) = g'(–1) =
f '(x)/x 0

1

3x  6x  6  8e 2x
2

1 1
 
0  0  6  8 14
1
9. Suppose f –1 is the inverse function of a differentiable function f and let G(x) = 1
.
f (x)
1 G (2)
If f(3) = 2 and f'(3) = , then the value of is DF0145
9 2
Ans. 0.50
1
Sol. G(x) = 1
f (x)

1
G '(x)  (f –1 (x))'
f (x) 
1 2

1
= 9
 f 1 (2)
2

40 E
 Methods of Differentiation

1
=  9  1
32

1
 f 1 (x)  
f '(x)

1

f '(3)

1
 9
1/ 9
f(3) = 2
f–1(2) = 3
d2y dy
10. If y = (A + Bx) emx + (m – 1)–2 ex and if 2
 2m  m 2 y  e kx , then the value of k is
dx dx
DF0153
Ans. 1
ex
Sol. y = (A + Bx)emx +
(m  1)

ex
y' = (A + Bx)emxm + Bemx +
(m  1)

ex
y" = (A + Bx)emxm2 + mBemx + Bemxm +
(m  1)

ex
 y" – 2my' + m2y = (A + Bx)m2emx + 2mBemx +
(m  1)

2me x m2ex
–2m2(a + Bx)emx – 2mBemx – +m 2
(A + Bx)e mx
+ 
(m  1)2 (m  1)2

(m  1)2 e x
  
(m  1)2

 = ex

So k = 1
cos x sin x cos x

11. Let f(x) = cos 2x sin 2x 2 cos 2x then the value of f    is
2
cos3x sin 3x 3cos3x
DF0151

E 41
 
JEE-Mathematics

Ans. 4
cos x sin x cos x
Sol. f(x) = cos 2x sin 2x 2 cos 2x
cos3x sin 3x 3cos3x

 sin x sin x cos x cos x cos x cos x cos x sin x  sin x

f '(x) =  sin 2x sin 2x 2 cos 2x  cos 2x 2 cos 2x 2 cos 2x  cos 2x sin 2x 4 sin 2x
3sin 3x sin 3x 3cos3x cos3x 3cos3x 3cos3x cos3x sin 3x 9sin 3x

1 1 0 0 1 0 0 1 1

f '    0 0 2  1 2 2  1 0 0
2
3 1 0 0 0 0 0 1 9


f '    4  0  8  4
2
logsin2 x cos x
12. lim has the value equal to
x 0 x
log 2  x  cos  
sin  
2
2
DF0150
Ans. 4
logsin2 x cos x
Sol. lim
cos  
x 0 x
log
sin 2  
x
2
2

loge sin 2  
x
loge cos x 2
 lim 
loge cos  
x 0 x loge (sin 2 x)
2

n sin 2  
x
n cos x 2
 lim  lim
n cos  
2
x 0 x x 0 n(sin x)
2

Use L-Hospital

42 E
 Methods of Differentiation

x 1
2 sin   cos   
x
2
(x / 2) 2 2 2

1 ( sin x) sin 2  
x x x
 x    
 lim cos x x  lim 2 2 2
x 0  x  x  0  1  sin x
  sin  1 
1
 2   x  x 2
 sin x   2
2 cos x  x
    x
 x 
2
cos   x
x 2 2
 
2 2

4×1=4

Aliter :

x
log sin 2
log cos x 2
lim 
x  0 log sin 2 x x
log cos
2

x
log sin 2
cos x  1 2
lim 
x 
cos2  1 log  4 sin 2 cos2 
x 0 x x
2  2 2

x
2 sin 2
2 1
lim
2 x
log  4 cos2 
x 0 x
2 sin
4
1  2
log  sin 2 
x
 2

x
lim 4 cos2 1  4
x 02 2
1
13. If the value of lim 1
is k then the value of [k2] is :-
x 0
(1  cosecx) n(sin x)

(where [.] denotes greatest integer function) DF0149

Ans. 7
1

Sol.  lim(1

 cosecx) nsin x
x 0

n(1  cosecx)
n  lim 
x 0 n sin x

Applying L'Hospital Rule

E 43
 
JEE-Mathematics

1
   cot x cosecx
(1  cosecx)
 n  lim
x 0 1
 cos x
sin x

cosecx
= lim
x 0 1  cosecx

1
= lim
x 0 1
1
cosecx

n  1

= e

n
  
 1  tan 2n 
14. If the value of Lim   is e/k then the value of k is
 1  sin 
n 

 3n 
DF0148
Ans. 6
n
  
 1  tan 2n 
Sol. lim  
 1  sin  
n 

 3n 

  
1 tan 2n 
lim  1n
n    
1sin
 3n 
 e 

  
 tan 2n sin 3n 
lim n
n  
1 sin
 e 3n


1
Let n =
t

   
 tan sin 3 t 
lim  2
t 0   
1 sin 3 t  t
 e

Applying L'Hospital Rule :

44 E
 Methods of Differentiation

 2   
sec 2 t. 2  cos 3 t. 3 
lim 
t 0     x
 1 sin t   t cos t.
  e  3  3 3


 
lim 
  e t 0 23


  e6 
15. Given a real valued function f(x) as follows :
x 2  2cosx  2 sin x  l n(e x cos x)
f(x) = for x < 0 ; f(0) = k & f(x) = for x > 0 and if f(x) is
x4 6x 2
continuous at x = 0, then the value of 3k is :-
DF0147
Ans. 0.25
Sol. For Continuity

L.H.L.

x 2  2 cos x  2
 lim
x 0 x4

2x  2 sin x
 lim
x 0 4x 3

2  2 cos x
 lim
x 0  12x 2

2sin x
 lim
x 0 24x

2 cos x
 lim
x 0 24

2 1
 
24 12

R.H.L.

sin x  n(e x cos x)

 lim
x 0 6x 2

1
cos x  x
 (e x cos x  e x ( sin x))
 lim e cos x
x 0  12x

E 45
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JEE-Mathematics

(cos x  sin x)
cos x 
 lim cos x
x 0 12x

cos x  1  tan x
 lim
x 0 12x

 sin x  0  sec2 x

12

1

12

R.H.L. = L.H.L.

1
f(0) =
12

So f(x) is continuous

1
3k  3   0.25
12
1  sin x  cosx  n(1  x)
16. The value of Lim 2
is 'a' then a2 is
x 0 x·tan x
DF0146
Ans. 0.25
1  sin x  cos x  n(1  x)
Sol. lim
x 0 tan 2 x
x  2  x2
x

1  sin x  cos x  n(1  x)

 lim
x 0 x3

1
cos x  sin x 
(1  x)
 lim  
x 0 3x3

1
 sin x  cos x 
(1  x)2
  lim 
x 0 6x

2
 cos x  sin x 
(1  x)3
  lim 
x 0 6

46 E
 Methods of Differentiation

1  0  2
  
6

3 1
   
6 2
x 6000  (sin x)6000
17. The value of Lim is DF0091
x 0 x 2 .(sin x)6000
Ans. 1000
x6000  (sin x)6000
Sol. lim
x 0 (sin x)6000
x 
2
6000
 x6000
x

x6000  (sin x)6000

 lim 
x 0 x 6002
6000

1  
sin x 

  lim  x2  
x 0 x

  Applying L'Hospital Rule

5999
x cos x  sin x
1  6000  
sin x 
 
   x  x2 
lim
x 0 2x

x cos x  sin x
  lim  3000  
x 0 x3

[cos x  sin x  cos x]

  lim  3000  
x 0 3x 2

3000 x sin x
  lim  2
x 0 3 x

 1000 
Paragraph for Question 18 to 20
Consider the function defined implicitly y2 + y – x = 2 on various intervals on the real line. If
 1
y   ,   , then equation implicitly defines a unique real valued differentiable function
 2
y = ƒ (x).
On the basis of above informations, answer the following questions.
18. The value of ƒ ''(4) is -
2 125 1 1
(A) (B) (C) (D)
125 2 5 125

E 47
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JEE-Mathematics

DF0165
Ans. A
dy dy
Sol. y2 + y – x = 2  2y  1
dx dx

dy 1
 
dx 2y  1

Put x = 4  y = 2, – 3

 1
y   ,    y = –3
 2

dy 1 1
   
dx x 4 2y  1 y 3 5

2
d2y  dy 
Now (1 + 2y) 2
 2   0
dx  dx 

d2 y 2 2
   f "(4)  
dx 2 x 4 125 125

19. If y = g(x) is inverse of y = ƒ (x), then g'(–2) is -

1
(A)  (B) –3 (C) –2 (D) 0
3
DF0166
Ans. B
Sol. g(x) is inverse of f(x)

 intercehanging x and y, we get

x2 + x – y = 2 where y = g(x)

 g'(x) = 2x + 1  g'(–2) = –3
20. If y = h(x) is mirror image of y = ƒ (x) about the line 2y + 1 = 0, then ƒ ''(x0) + h''(x0), where
 9 
x0    ,   is -
 4 
2 2
(A) 2x0 (B) (C) (D) 0 DF0167
2ƒ(x 0 )  1 (2ƒ(x 0 )  1)3
Ans. D
1
Sol. h(x) is mirror image of f(x) in y =
2

f(x)  h(x) 1
   f(x) + h(x) = –1
2 2

48 E
 Methods of Differentiation

 9 
 f"(x) + h"(x) = 0  x    ,  
 4 

EXERCISE # (JM)

 tan   cot   1  3  dy 5
1. If y()  2    2 ,   ,   , then at   is :
 1  tan   sin 
2
 4  d 6
[JEE(Main)-2020]
1 4
(1) 4 (2)  (3) (4) –4
4 3
DF0155
Ans. 1
 tan   cot   1  3 
Sol. y()  2    2 ,  , 
 1  tan   sin 
2
 4 

| sin   cos  | (sin   cos )

 
| sin  | sin 

= –1 – cot

y'() = cosec2

 5 
y '   4
 6 

6
3 4  dy
2. If y   k cos
k 1
1
 cos kx  sin kx  , then
5 5  dx
at x = 0 is________. [JEE(Main)-2020]

DF0156
Ans. 91
3 4 
Sol. Put cos  ,sin   ,0   
5 5 2
3 4 
Now cos  ,sin   ,0   
5 5 2
3 4
Now coskx – sinkx
5 5
= cos.coskx – sin.sinkx
= cos( + kx)
As we have to find derivate at x = 0
We have cos–1(cos( + kx))
= ( + kx)
6
 y   (  kx)
k 1

E 49
 
JEE-Mathematics

dy 6
6  7  13
  k   91
dx at x 0 k 1 6
 P x
 ; x2
3. Consider the function f  x    sin  x  2 
 7 ; x2

Where P(x) is a polynomial such that P"(x) is always a constant and P(3) = 9. If f(x) is continuous at
x = 2, then P(5) is equal to ______. [JEE(Main)-2021]
DF0157
Ans. 39
 P(x)
 , x2
Sol. f '(x)   sin(x  2)
 7, x2

P"(x) = const.  P(x) is atmost 2nd degree polynomial f(x) is cont. at x = 2
f(x) is cont. at x = 1
f(2+) = f(2–) = f(2)
P(x)
lim 7
x 2 sin(x  2)

(x  2)(ax  b)
lim  7  2a + b = 7
x 2 sin(x  2)
P(x) = (x – 2)(ax + b)
P(3) = (3 – 2)(3a + b) = 9  3a  b  9
a  2,b  3
P(5) = (5 – 2)(2.5 + 3) = 3.13 = 39
  1 x 
4. Let f(x) = cos  2tan 1 sin  cot 1   , 0 < x < 1. Then : [JEE(Main)-2021]
  x 
(1) (1 – x)2 f'(x) – 2(f(x))2 = 0 (2) (1 + x)2 f'(x) + 2(f(x))2 = 0
(3) (1 – x)2 f'(x) + 2(f(x))2 = 0 (4) (1 + x)2 f'(x) – 2(f(x))2 = 0
DF0158
Ans. 3
  
Sol. f(x)  cos  2 tan 1 sin  cot 1 1  x   (0 < x < 1)
  x 

1 x
cot 1  sin 1 x
x

 f(x) = cos(2tan–1 x )
  
= cos  tan 1  2 x  
  1 x 

50 E
 Methods of Differentiation

1 x
f(x) =
1 x
2
Now f '(x) =
(1  x)2
2
2  1 x 
 f ' (x) (1 – x) = –2  
 1 x 
 (1 – x)2 f '(x) + 2(f(x))2 = 0
5. Let f (x) be a polynomial function such that f (x) + f ' (x) + f ''(x) = x5 + 64. Then, the value of
f(x)
lim
x 1 x  1

(A) –15 (B) –60 (C) 60 (D) 15

[JEE(Main)-2022]
DF0168
Ans. A
f(x)
Sol. lim  f '(1) (and f(1) = 0)
x 1 x 1
f(x) + f '(x) + f "(x) = x5 + 64
f ' (x) + f "(x) + f '"(x) = 5x4
f " (x) + f '"(x) + f iv (x) = 20x3
f '" (x) + f iv (x) + f v (x) = 60x2
 f v (x) – f '' (x) = 60x2 – 20x3
 120 – f''(1) = 40  f "(1) = 80
Also f(1) + f '(1) + f "(1) = 65  f ' (1) = –15.
f '(4)
6. Let f :  satisfy f(x + y) = 2xf(y) + 4yf(x),  x, y  . If f(2) = 3, then 14. is equal to
f '(2)
_____.
[JEE(Main)-2022]
DF0169
Ans. 248
Sol. Put y = 2

f(x + y) = 2x.f(y) + 4y.f(x)

f(x + 2) = 2x . 3 + 16 f(x)

f ' (x + 2) = 16 f '(x) + 3.2x ln2

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JEE-Mathematics

f ' (4) = 16 f '(2) + 12ln2 …….(1)

f(y + 2) = 4f(y) + 3.4y

f '(y + 2) = 4f '(y) + 3.4yln4

f ' (4) = 4f ' (2) + 96 ln2 ……(ii)

solving equation (i) and (ii), we get

f '(2) = 7 ln2

from equation (i), we get

f '(4) = 124 ln2

f '(4)
Now,  14.
f '(2)

124 ln 2
14 
7 ln 2

= 248

a 1 0
7. Let f(x)  ax a 1 , a  R. Then the sum of the squares of all the values of a for which
ax 2 ax a
2f (10)  f (5)  100  0 is :
(A) 117 (B) 106 (C) 125 (D) 136
[JEE(Main)-2022]
DF0170
Ans. C
a 1 0
Sol. f(x)  ax a 1
ax 2 ax a

1 1 0
f(x)  a x a 1
2
x ax a

52 E
 Methods of Differentiation

= a[1(a2 + ax) + 1(ax + x2)]
 f(x) = a(x + a)2
so, f '(x) = a(x + a)2
as, 2f '(10) – f ' (5) + 100 = 0
 2 × 2a(10 + a) – 2a(5 + a) + 100 = 0
 40a + 4a2 – 10a – 2a2 + 100 = 0
2a2 + 30a + 100 = 0
 a2 + 15a + 50 = 0
(a + 10)(a + 5) = 0
a = –10 or a = –5
Required = (–10)2 + (–5)2 = 125
    3

8. Let y  f  x   sin 3   cos 
  
4x 3  5x 2  1 2    . Then, at x = 1,

3 3 2   

(1) 2y' 32 y  0 (2) 2y ' 32 y  0

(3) 2y' 32 y  0 (4) y ' 32 y  0

[JEE(Main)-2023]
Official Ans. by NTA (2)
Allen Ans. (2)
Sol. y = sin3(/3cosg(x))

gx   4x 
3/2
3
 5x 2  1
3 2
g(1) = 2/3
   
y '  3sin 2  cos g  x    cos  cos g  x  
3  3 


3
 sin g  x   g'  x 
     2 
y ' 1  3sin 2    .cos   .   sin g' 1
 6 6 3 3 


g'  x    4x   12x 
1/2
3
 5x 2  1 2
 10x
3 2

g' 1 
2 2

 2   2   
3 3  3 32
y' 1  . .      
4 2 3  2  16

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JEE-Mathematics

1
y 1  sin 3   / 3cos2  / 3   
8
2y' 1  32 y 1  0
9. If f(x) = x2 + g(1)x + g(2) and g(x) = f(1)x2 + xf(x) + f(x), then the value of f(4) – g(4) is
equal to _______.
[JEE(Main)-2023]
Official Ans. by NTA (14)
Allen Ans. (14)
Sol. f(x) = x2 + g(1)x + g (2)
f (x) = 2x + g(1)
f(x) = 2
g(x) = f(1) x2 + x [2x + g(1)] + 2
g(x) = 2f(1) x + 4x + g(1)
g(x) = 2f(1) + 4
g(x) = 0
2f(1) + 4 = 0
f(1) = – 2
–2 = 1 + g(1) = g(1) = – 3
So, f’(x) = 2x – 3
2
f(x) = x – 3x + c
c=0
f(x) = x2 – 3x
g(x) = – 3x + 2
f(4) – g(4) = 14
1 1 1
10. For the differentiable function f :  – {0} , let 3f(x)  2f     10 , then f (3)  f '   is
x x 4
equal to
33 29
(1) 7 (2) (3) (4) 13
5 5
[JEE(Main)-2023]
Official Ans. by NTA (4)
Allen Ans. (4)
 1 1 
Sol. 3f (x)  2f     10   3
  
x x 
 1 
 2f (x)  3f  x   x  10   2
   

54 E
 Methods of Differentiation

3
5f (x)   2x  10
x
1 3 
f (x)    2x  10 
5 x 
1 3 
f '(x)    2  2 
5 x 

1 1 1
f (3)  f '    (1  6  10)  (48  2)
4 5 5
= |–3–10| = 13

11. Suppose

2 x
 
 2  x tan x tan 1 x 2  x  1 
f x  ,
 7x  3x  1
3
2

Then the value of f '(0) is equal to


(1)  (2) 0 (3)  (4)
2
[JEE(Main)-2024]
Ans. (3)
f(h)  f(0)
Sol. f '(0)  lim
h 0 h

(2h  2h ) tan h tan 1(h 2  h  1)  0

= lim
h0 (7h2  3h  1)3 h

= 
12. Let ƒ(x) = x5 + 2ex/4 for all x  R. Consider a function g(x) such that (gof) (x) = x for all x  R.
Then the value of 8g(2) is :
(1) 16 (2) 4 (3) 8 (4) 2
[JEE(Main)-2024]
Ans. (1)
Sol. f(x) = 2
when x = 0
 g(f(x)) f(x) = 1

1
g  2  
f 0
 

E 55
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JEE-Mathematics

2 x/4
 f(x) = 5x4 + e
4

g(2) = 2
Ans = 16
Option (1)

EXERCISE (JA)
1. Let ƒ :   , g :    and h :    be differentiable functions such that ƒ(x) = x3 + 3x + 2,

g(ƒ(x)) = x and h(g(g(x))) = x for all x  . Then- [JEE(Advanced)-2016, 4(–2)]

1
(A) g '  2   (B) h'(1) = 666 (C) h(0) = 16 (D) h(g(3)) = 36
15
DF0113
Ans. B,C
Sol. Given
g(f(x)) = x
 g'(f(x)).f ' (x) = 1
1
 g'(f(x)) =
f '(x)
1
 g'(x3 + 3x + 2) =
3x  3
2

Put x = 0

1
 g'(2) 
3
option A is wrong
* h(g(g(x))) = x ; given
x 
replace
 f(x)

 h(g(g(f(x)))) = f(x) ; given g(f(x))  x

 h(g)(x)  f(x) 

Put
x 3
h(g(3)) = f(3) = 33 + 3 × 3 + 2  h(g(3))  38

 x 
Replace
f(x)

 h (g(f(x)))  f(f(x))
x

 h(x)  f((x)) 

Put
x 0
h(0)  f (f(0))  f(2)  16  h(0)  16
2

56 E
 Methods of Differentiation

 h'(x) = f '(f(x)).f '(x)
 h'(x) = f '(x3 + 3x + 2).(3x2 + 3)
Put x = 1
 h'(1) = f '(6).6
 h'(1) = f '(6).6
 h'(1) = 111 × 6  h'(1) = 666

E 57
 
JEE-Mathematics

ANSWER KEY
Do yourself -1

1 1
(i) (a) (b) 
x x2

Do yourself -2

(i) (a) 3x2 + 12x + 11 (b) 5e5x tan (x2 + 2) + 2xe5x sec2(x2 + 2)

Do yourself -3

(i) xx (nx + 1) (ii) y(1 + 2x + 3x2 + 4x3)

Do yourself -4

1
(i) –1 (ii) (iii) 2(xnx)(nx)
t

Do yourself -5

cos(x  y)  1  2x  e y 
(i) (ii) y '    y  , –1
cos(x  y)  1  xe  1 

Do yourself -6

(i) (a) –6 (b) 3 (c) –6

Do yourself -7

(i) D

Do yourself -8

(i) y'' = 4y + 2xy' (ii) +4

Do yourself -9

(i) e( sin 1 + cos 1) – 1 (ii) 9

Do yourself -10

1 1
(i) (a) (b)
3 2

58 E
 Methods of Differentiation

Elementary Problems on Differentiation
sin x  cos x  x(sin x  cos x) sin x
1. 2. tan4x 3. 2x
1  sin 2x cos3 x
a x 1 1
4.  sin 5. 6.
3 3 x 1
2 1  2 tan x.cos2 x
2 cos
2
1 x cos 1  x 2
7. –12cos24x sin4x 8. 9.
x x 1  x2
4 tan .cos2
2 2
2x
10.  11. 4(1 + sin2x)3sin2x
3sin 2 3
1  x . (1  x )
2 3 2 2

1 x 
sin 2  
12.
x2  1
13. 1 x  14. –3sin3x sin(2cos3x)
 1   1  x(1  x)2
2x 2 cos2  x   1  tan  x  
 x  x
x  2 arcsin x
15. arcsin x  16. 17.
1  x2 2(arccos x) 2
1 x 2
1  x2
1 x sin x
18.  19. sin x.arctan x  x cos x.arctan x 
(arcsin x)2 1  x 2 1  x2
2x 2 2x
20. 0 21.  22.
(1  x 2 )2 1  x4
x 1 a 2  b2
23. 24.
8 4 (arcsin x 2  2x)3 (1  2x  x 2 )(x 2  2x) a  b cos x
x 1
25. 2x log3 x  26.
n3 2x nx
1  n nx 2
27. sin x nx  x cos x nx  sin x 28. n 1
29.
x x(1  nx) 2
2x  4 2
30. 31.
x  4x
2
sin 2x
1 x
32. 33.
x log 5 x log3 (log 5 x) n2 n3 n5 arctan 1  x 2 (2  x 2 ) 1  x 2
x3
cot
4 2x ( n2  1)  3x 2  x 3
34. 35. 10x (1 + x n10) 36.
3 2 x3 ex
12 n sin
4
sin x  cos x ( nx  1) n2 nx x
37. – 38. 2
ex n2x
2.10 x n10 2 2
39.  40. 3sin x cos x. n3 41. 2ab2 xe  b x
(1  10 )x 2

E 59
 
JEE-Mathematics
2
42. Ae  k x [  cos(x  )  k 2 sin(x  )]

x  1  cos2 x 
43. x x .x x  n 2 x  nx   44. (sin x)cosx   sin x n sin x 
 x  sin x 

 1  57x 2  302x  361 (x  1) 2 4 x  2

45. ( nx) x   n nx  46. 2x nx 1 nx 47. .
 nx  20(x  2)(x  3) 5
(x  3) 2

1 1 1 ex  1
2
48. x sin x 1  e x   cot x  .  49. x x (1  nx)
2 x 2 1  ex 
1
 2x sin x 
 
x
50. x 2
(2  nx) 51. (x 2  1)sin x  2  cos x n x 2  1 
 x 1 

x 4  6x 2  1 x(x 2  1)
52. 3
3x(1  x 4 ) (x 2  1)2

EXERCISE # O-1
Que. 1 2 3 4 5 6 7 8 9 10
Ans. A B A A B C B D A B
Que. 11 12 13 14 15 16 17 18 19 20
Ans. A C B D A C A D B A

EXERCISE # O-2
Que. 1 2 3 4 5 6 7 8 9 10
Ans. A D B D C D A B,D C,D A,C,D

EXERCISE # O-3
Que. 1 2 3 4 5 6 7 8 9 10
Ans. 16 5 12 25 6 0 2 2 0.50 1
Que. 11 12 13 14 15 16 17 18 19 20
Ans. 4 4 7 6 0.25 0.25 1000 A B D

EXERCISE # JEE-MAIN
Que. 1 2 3 4 5 6 7 8 9 10
Ans. 1 91 39 3 A 248 C 2 14 4
Que. 11 12
Ans. 3 1

EXERCISE # JEE-ADVANCED
Que. 1
Ans. B,C

60 E