DIFFERENTIATION

SYLLABUS
Differentiability of a function at a given point, Differentiation of the sum, difference,
product and quotient of two functions, differentiation of trigonometric functions, inverse
trigonometric functions, logarithmic functions, exponential functions, composite functions
and implicit functions. derivatives of order upto two.

DIFFERENTIABILITY OF A FUNCTION AT A GIVEN POINT

Let f (x) be a real valued function defined on an open interval (a,b) where c
 ( a,b ) . Then f (x) is said to be differentiable or derivable at x = c,
f(x) − f(c)
iff lim exists finitely.
x →c (x − c)
This limit is called the derivative or differential coefficient of the function
d
f(x) at x = c, and is denoted by f’(c) or D f (c) or (f (x))x = c
dx
 Thus, f (x) is differentiable at x = c
f(x) − f(c)
lim exists finitely
x →c (x − c)
f(x) − f(c) f(x) − f(c)
 lim− = lim+
x →c (x − c) x →c (x − c)
f(c - h) - f(c) f(c + h) − f(c)
Þ lim = lim
h® 0 ( - h) h → 0 (h)
f(x) − f(c) f(c − h) − f(c)
Here lim− = lim is called the left hand derivative of f (x) at x = c
x →c (x − c) h → 0 ( −h)
and is denoted by f’ (c − ) or LF’(c).
f(x) − f(c) f(c + h) − f(c)
While , lim+ = lim is called the right hand derivative
x →c (x − c) h → 0 (h)
of f (x) at x=c and is denoted by f’ (c + ) or Rf’ (c).
Thus f (x) is differentiable at x = c.
Lf’(c) = Rf’ (c)
If Lf’ (c)  Rf’(c) we say that f (x) is not differentiable at x = c.

Example -1: The set of triplets (a, b, c) of real numbers with a 0, for which the
function
 x for x  1
f(x) =  2 , is differentiable, is
ax + bx + c otherwise
(A) { ( a, 1- 2a, a) / a R; a  0 }
(B) { ( a, 1- 2a, c) /a, c R; a  0 }
(C) { (a, b, c)/ a, b, c R; a + b+ c = 1 }
 (D) { ( a, 1- 2a, 0) / a R; a  0 }

Solution: (A) Given f is differentiable for all real x

 f is continuous for all real x.
so, lim f(x) = f(1)  a + b + c =1 . . . . (1)
x →1


 1 for x  1
Also f(x) = 
2ax + b otherwise (sin ce f is continuous)

f (1+) = f(1–)  1 = 2a + b  b = – 2a + 1 . . . . (2)
since a, b, c  R and a  0, using (1) and (2)
 c=a

DIFFERENTIABILITY IN AN INTERVAL
2.1 IN OPEN INTERVAL
A function f(x) defined on an open interval (a, b) is said to be differentiable or derivable
in open interval (a, b) if it is differentiable at each point of (a, b)

2.2 IN CLOSE INTERVAL

A function f(x) defined on [a, b] is said to be differentiable or derivable at the end points a
and b if it is differentiable from the right at a and from the left at b. In other words
f(x) − f(a) f(x) − f(b)
lim+ and lim+ both exist.
x →a x−a x →b x −b
“If f is derivable in the open interval (a, b) and also at the end points a and b, then f is
said to be derivable in the closed interval [a, b]”.
For checking differentiability on a closed interval [a, b] we say,
“A function f is said to be differentiable function if it is differentiable at every point of its
domain.”

Example -2: Let f(x) = x3 – x2 + x + 1

max f (t ), 0  t  x, forx  1
g(x) = 
3 − x, 1 x  2
Discuss the continuity and differentiability of g(x) in (0, 2).
Solution: f(x) = x3 – x2 + x + 1
 f(x) = 3x2 – 2x + 1 > 0  x
 f(x) is an increasing function on [0, x]
x 3 − x 2 + x + 1, 0  x  1
Hence, g(x) = 
3 − x, 1 x  2
Clearly g(x) is continuous at x = 1 and not differentiable at x = 1.

DERIVATIVE OF F(X) FROM THE FIRST PRINCIPLE

dy f ( x + h) − f ( x )
= lim = f (x)
dx h →0 h
The limit which is a function of x is called the derivative of f (x) and is denoted by f (x).
The symbol f (c), then denotes the value of the function f (x) for x = c.

Example -3: Find the derivative of x3.

Solution: Let f (x) = x3

f ( x + h) − f ( x ) ( x + h)3 − x 3
So f (x) = lim = lim
h→0 h h →0 h
2 2
= lim h + 3 xh + 3 x
h→0

dy
= f (x) = 3x2.
dx

1
Example -4: Find the derivative of .
x
1
Solution: Let f (x) =
x
1 1
−
dy f ( x + h) − f ( x )x+h x
So, = lim = lim
dx h → 0 h h
h→0

x − x+h x + x +h −1
= lim  = 3/2
h →0 x ( x + h )h x + x + h 2x

LIST OF DERIVATIVES OF IMPORTANT FUNCTIONS

d n d
1. (x ) = nx n−1 7. (tan x) = sec 2 x
dx dx
d d
2. (x) = 1 8. (cot x) = − cosec 2 x
dx dx
d  1 1 d
3.   =− 2 9. (sec x) = sec x tan x
dx  x  x dx
d  1 n d
4.  n 
= − n+1 ,xL 0 10. (cose x) = − cosec x cot x
dx  x  x dx
d d x
5. (sin x) = cos x 11. (e ) = e x
dx dx
d d 1
6. (cos x) = − sin x 12. (Inx) =
dx dx x
5
Example -5: Find the derivative of x 2 .
 d (x )5 / 2 5
Solution: = x3 / 2
dx 2

1
Example -6: Find the derivative of (1 + x )2 .
d (1 + x )1 / 2 1
Solution: = (1 + x )−1 / 2
dx 2

Example -7: Find the derivative of xe .

Solution:
( )
d xe
= ex e −1
dx

FUNDAMENTAL RULES FOR DIFFERENTIATION

(i) Differentiation of a constant function = 0
d
i.e. C= 0
dx
(ii) Derivative of the sum: Let u, v be two derivable functions of x so denoting their
sum by y, we write, y = u + v
dy du dv
so, = +
dx dx dx
(iii). Derivative of the difference: Let u, v be two derivable function of x, so denoting
their difference by y, so we write y = u –v
dy du dv
so, = −
dx dx dx

(iv). Generalisation: By a repeated application of the results obtained above, it can be

proved that if u1, u2 ……, un be any finite number of derivable functions, then
y = u1  u2  u3  …….  un
dy du1 du2 du3 du
So, =    .......  n
dx dx dx dx dx
dy du du du
 = (u2u3 ....un ) 1 + (u1u3 ....un ) 2 + .... + (u1u2 ....un −1 ) n
dx dx dx dx

dy x2 + 1
Example -8: Find , if y = .
dx x

1 1
Solution: Let y = x + , so let u = x1 and v =
x x
So y = u + v
dy du dv 1 x2 − 1
= − =1– 2 =
dx dx dx x x2
 dy x3 - 1
Example -9: Find , if y = .
dx x

1 1
Solution: Let y = x2 – , so let u = x2 and v =
x x
So y = u – v
dy du dv 1 2x 2 + 1
= − = 2x + 2 =
dx dx dx x x2

dy
Example -10: Find of
dx
3x + 4
(a). y = (x + 2)(x + 3) (b). y =
4x + 5
(c). y = 1 + x2 (d). y = ax 2 + 2 bx + c

dy
Solution: (a). = (x + 2).1 + (x + 3).1
dx
= x + 2 + x + 3 = 2x + 5
(b).
dy
=
(3x + 4 )x − 1 4 + 3( 4x + 5) = − 1
dx (4x + 5 )2 ( 4 x + 5 )2 ( 4 x + 5 )2

(c).
dy 1
(
= 1 + x2
dx 2
)
−1 / 2
 2x =
x
1 + x2

(d).
dy 1
(
= ax 2 + 2bx + c
dx 2
)−1 / 2
(2ax + 2b )
= (ax 2 + 2bx + c ) (ax + b)
−1/ 2

DERIVATIVES OF FUNCTIONS IN PARAMETRIC FORMS

Sometimes the relation between two variables is neither explicit nor implicit, but some
link of a third variable with each of the first two variables, separately, establishes a
relation between the first two variables. In such a situation, we say that the relation
between them is expressed parametrically. The third variable is called the parameter.
More precisely, a relation expressed between two variables x and y in the following form
x = f (t), y = g(t) is said to be parametric form with ‘t’ as a parameter. In order to find
differentiation of functions in such form, we have by chain rule.
dy dy dx
= .
dt dx dt
 dy
dy dt  dx 
or =  whenever dt  0 
dx dx  
dt
dy g'(t)  dy dx 
Thus =  as dt = g'(t)and dt = f '(t) 
dx f '(t)  
providedf '(t)  0
dy
Example -11: Find of x = 2at2 , y = 2at3
dx
dx dy
Solution: = 4at , = 6at 2
dt dt
dy dy dt 6at 2 3t
= . = =
dx dt dx 4at 2

2 dy
Example -12: If x= e − t and y = tan–1 (2t + 1), find .
dx
2
Solution: Here x= e − t
On differentiating both sides, we get,
dx 2
 = e− t .( −2t)
dt
and y = tan–1 (2t + 1)
On differentiating both sides, we get,
dy 1
 = (2)
dt 1 + (2t + 1)2
dy 2
dy dt 1 + (4t + 4t + 1)
2
 = =
dt dx 2t
− t2
dt e
2
dy −e t
Hence, =
dt 2t(2t 2 + 2t + 1)

DERIVATIVES OF A COMPOSITE FUNCTION

Given a composite function y = f(x), i.e. a function represented by y = F(u), u =  (x) or y
dy dF du
= F [  (x)], then y ' = = .
dx du dx
This is called the chain rule. The rule can be extended to any number of composite
variables;
dy df du dv
e.g., if y = f(u(v)), then y ' = = . .
dx du dv dx
Example -13: Find dy/dx
 x 
(i) sin cos x (ii) sin  log 
 x + 1 

(iii) (
log x + e x
)
Solution:
(i)
(
dy d sin cos x d cos x d ( cos x )
= . .
) ( )
dx d cos x ( d ( cos x ) )
dx

1 sin x.cos x cos x

= cos cos x. .( − sin x) = −
2 cos x 2 cos x
 x 
(ii) Let y = sin  log 
 x+1 

dy  x  d  x 
= cos  log   log 
dx  x + 1  dx  x + 1 

 x 1 d
= cos  log
 x + 1

2 dx
log x − log(x + 1)
 
1  x 1 1 
= cos  log   −
2 
 x + 1   x x + 1

1  x  1
= cos  log 
2  x + 1  x(x + 1)

(iii) (
y = log x + e x
)
=
x+e
1
x

1 + e

x d
dx
( x ) = = x +1e x

1 + e

x 1 

2 x
2 x +e x
=
2 x x+e ( X
)
DERIVATIVE OF AN IMPLICIT FUNCTION
If x and y are related by the rule F (x, y) = 0 such that y cannot be obtained entirely or
exactly in terms of x then y is said to be an implicit function of x.
For example: x 2 + y 2 = r 2  y =  a2 − x 2
Here we do not get a unique value of y for each x.
Eg. x2 – y3 + 3x2y = 0
 dy
Here also y cannot be obtained entirely in terms of x. To find in such cases start
dx
differentiating the given equation as it is (using rule of composite functions)
dy dy −3x 2
For example: for x3 + y2 = a2, we have 3x2 + 2y =0  =
dx dx 2y
If y can be expressed entirely in terms of x, then y is said to be an explicit function of x.
Note that every explicit function can be written as the implicit function y – f(x) = 0.

Example -14: Find dy/dx

(i) log(xy) = x2 + y2 (ii) x + y = sin (xy)
Solution: (i) log (xy) = x + y  logx + logy = x2 + y2
2 2

Differentiating w.r.t x
1 1 dy dy 1  dy 1
+ = 2x = 2y   − 2y  = 2x −
x y dx dx y  dx x
1 − 2y 2 dy 2x 2 − 1 dy y(2x 2 − 1)
=  =
y dx x dx x(1 − 2y 2 )
(ii) x + y = sin (xy)
dy  dy 
Differentiating w.r.t. x, we get 1 + = cos (xy).  x + y
dx  dx 
dy
 1 − x cos(xy) = y cos(xy) − 1
dx
dy y cos(xy) − 1
 =
dx 1 − x cos(xy)

INVERSE FUNCTIONS AND THEIR DERIVATIVES

Theorem: If the inverse functions f and g are defined by y = f(x) and x = g(y) and if
1
f’(x) exists and f’(x)  0 then g’(y) = . This result can also be written
f '(x)
dy dy dx 1 dy dx
as, if exists and  0 , then = or . =1 or
dx dx dy  dy  dx dy
 dx 
 
dy 1  dx 
=   0
dx  dx   dy 
 dy 
 
Result:
1 −1
1. D(sin−1 x) = , − 1 x  1 2. D(cos−1 x) = , − 1 x  1
1− x 2
1− x2
1 1
3. D(tan−1 x) = , x R 4. D(sec −1 x) = , x 1
1+ x2 x x2 − 1
 −1 −1
5. D(cosec −1x) = , x 1 6. D(cot −1 x) = , x R
x x2 − 1 1+ x2

dy
Example -15: Find of
dx
(a). y = sin 2x (b). y = cos 2x
(c). y = x sin–1 x (d). y = a tan–1 x
(e). y = sec2 x (f). y = (log2 x)2
dy
Solution: (a). = 2 cos 2x
dx
dy
(b). = – 2 sin 2x
dx
dy 1
(c). = x. + sin–1 x
dx 1− x 2

dy 1
(d). =a
dx 1 + x2
dy
(e). = 2 sec x . sec x tan x = 2 sec2 x tan x
dx
dy 1 1
(f). = 2 log2 x  loge2 = 2log2x . log2e
dx x x

HIGHER ORDER DERIVATIVES

Let y = f(x)
dy
First derivative = f ' (x) = y1
dx
d2 y
Second derivative 2 = f '' (x) = y 2
dx
3
dy
Third derivative 3 = f ''' (x) = y 3 etc.
dx

Example -16: Find the second derivative of ax3 + bx2 + cx + d.

Solution : Let y = ax3 + bx2 + cx + d.
dy
 = 3ax 2 + 2bx + c
dx
d  dy 
 = 3a(2x) + 2b (1) + 0
dx  dx 
d2 y
 2 = 6ax + 2b
dx

Logarithmic Differentiation
If u and v are functions of independent variable x then to differentiate the
functions like uv , first we take the log and then differentiate.

Example -17: Differentiate (log x )tan x w.r.t. sin( m cos– 1x).

Let y = (log x )
tan x
Solution: and z = sin(m cos– 1x)
So logy = tanx log log x
1 dy 1 1
. = sec 2 x. log log x + tan x  
y dx log x x
dy tan x  tan x 
= (log x )  sec 2 x. log log x + 
dx  x log x 

and
dz
(
= cos m cos −1 x  m  )
−1
=
(
− m cos m cos −1 x )
dx 1 − x2 1 − x2
tan x  tan x 
− (log x )  sec 2 x . log log x +  1 − x 2
dy / dx  x log x 
=
so
dz / dx (
m cos m cos −1 x ) .

dy
Example -18: If xy . yx = 1, find .
dx
Solution: Taking log on both sides;
Y log x +x log y = log 1
Differentiating both sides, we get,
d d  d  d 
y. (log x) +  y  .log x +  x  .log y + x  log y  = o
dx  dx   dx   dx 
1 dy 1 dy
or y. + log x. + 1.log y + x. . =0
x dx y dx
 x  dy y 
or log x. + y  + dx = −  x + log y 
   
dy (y + x log y) y
=− .
dx (x + y log x) x

Differentiating a function w.r.t. to another function

Let we have to differentiate f(x) with respect to g(x). If y = f(x) and t = g(x) then we have
dy dy dt dy dy dt
to find . First we find and then = ¸
dx dx dx dt dx dx

Example -19: Differentiate sin2x w. r. t. (logx)2

Solution: Let y = sin2x and t = (logx)2
 dy dt  1  2log x
2sin x cos x, = 2log x   =
dx dx  x x
dy dy dt 2sin x cos x x sin x cos x
= ¸ = .x =
dt dx dx 2log x log x

dy
Example -20: If x2 + y2 + xy = 2, find .
dx
Solution: x2 + y2 + xy = 2,
Differentiating both sides we get,
d 2 d 2 d d
dx
( )
x +
dx
( )
y +
dx
(xy) =
dx
(2)

dy  dx  d 
or 2x + 2y +   y + x  y = 0
dx  dx   dx 
dy dy
or 2x + 2y + 1.y + x. =0
dx dx
dy
or (2y + x) = −(2x + y)
dx
dy (2x + y)
or =−
dx (2y + x)

dy
Example -21: If y = sin x + sin x + sin x + ...., find .
dx
Solution: y = sin x + sin x + sin x + ....,
 y= ( sin x ) + y
 y 2 = sin x + y
Differentiating both sides, we get
dy dy
2y = cos x +
dx dx
dy
 ( 2y − 1) = cos x
dx
dy cos x
 =
dx 2y − 1
  1 dy
Example -22: If 5f(x) + 3f   = x + 2 and y = xf (x), then find at x = 1.
x dx
 1
Solution: Here, 5f(x) + 3f   = x + 2
x
1
Put x = we get,
x
 1 1
5f   + 3f(x) = + 2
x x
Solving (i) and (ii) we get,
3
16f(x) = 5x − + 4
x
 y = x f (x)
1  3 
 y = x. 5x − = 4 
16  x 
1
y=
16
 
5x 2 − 3 + 4
dy 1
or = 10x + 4
dx 16
dy
Now at x = 1
dx
 dy  10 + 4 14 7
 dx  = = =
 at x =1 16 16 8
 dy  7
   =
 dx at x =1 8

Example -23: Find a, b, c and d, where f (x) = (ax + b) cos x + (cx + d) sin x and f’
(x) = x cos x is identity in x.
Solution: Here,
f’ (x) = x cos x
 a cos x −(ax + b) sin x + c sin x + (cx + d) cos x  x cos x
or (a + cx + d) cos x + (−ax – b + c) sin x  x cos x + 0.sin x
 a + b + cx = x and −ax−b + c = 0
Which is again identity in ‘x’
 a + b = 0, c = 1, −a = o, −b + c = o
 a =0, b = 1, c = 1, d = 0

SOLVED PROBLEMS
 SUBJECTIVE

1. The function f(x) = (x2 – 1) |x2 – 3x + 2| + cos (|x|) is not differentiable at :

(A) x = –1 (B) x=0
(C) x=1 (D) x=2
Solution : (D) Since cos (–x) = cos x
 cos | x | is differentiable for each x Î R .
Also, x2 − 3x + 2  0  ( x − 2)( x − 1)  0
 x ( −, 1)  ( 2, )
Similarly, x2 − 3x + 2  0  x (1, 2)
If g (x ) = (x 2 - 1) | x 2 - 3x + 2 | , then f (x) is not differentiable at points where g(x)
is so.
Now, g ( x) = ( x − 1) ( x + 1)( x − 2) ,  x  ( −, 1)  ( 2, )
2

= − ( x − 1) ( x + 1)( x − 2)  x  (1, 2)
2
and
 g (x) is not differentiable at x = 2
Þ f (x) is not differentiable at x = 2.
Hence (D) is correct answer.

y = 2− log 2 (x ) , then dy is
3
−5
2.
dx
− 3x 2 x2
(A) (B)
(x 3
−5 )
2
(x 3
−5 )
2

3x 2
(C) (D) none of these
(x 3
−5 ) 2

1 3
− 5 ))
Solution: y = 2log 2 (1 /( x = (x3 − 5)−1= 3
x −5
dy − 3x 2
= − 1 (x3 − 5)−2  3x2 = 3
dx ( x − 5 )2
Hence (A) is correct answer.
dy
3. y = esin 2x, then is
dx
(A) x cos 2x (B) 2y cos 2x
(C) xy cos 2x (D) none of these
dy
Solution: y = esin2x  = esin2x cos 2x . 2 = 2 esin2x. cos 2x
dx
Hence (B) is correct answer.

dy
4. y = (sinx + cosx)x , then is
dx
  log y x(cos x − sin x )
(A) y(sinx + cosx) (B) y  + 
 x sin x + cos x 
 log y 
(C) y + y2  (D) none of these
 x 
Solution: y = (sin x + cos x)x
log y = x log (sin x + cos x)
1 dy x  (cos x − sin x )
 = log (sin x + cos x) +
y dx sin x + cos x
dy  x(cos x − sin x 
= (sin x + cos x)x  + log(sin x + cos x )
dx  sin x + cos x 
Hence (B) is correct answer.

y = elog e (sin ) , then dy is

−1
x + cos −1 x
5.
dx
(A) 5 (B) 2
(C) 0 (D) none of these
−1
x + cos −1 x )  dy
Solution: y = elog e (sin = sin−1 x + cos−1 x =  =0
2 dx
Hence (C) is correct answer.

−1 −1
( x −2) dy
6. y = 2sin x
 ecos , then is
dx
y log 2 y log(2 / e )
(A) (B)
1 − x2 1− x2
y −1
(C) (D) none of these
1− x2
−1 −1
Solution: y = 2sin x  ecos ( x − 2)
log y = sin−1 x log 2 + cos−1 (x − 2)
1 dy log 2 1 log 2 1
 = − = −
y dx 1− x 2
1 − ( x − 2)2
1− x 2 2
1 − x − 4 + 4x
= answer is none of these.
Hence (D) is correct answer.

dy
7. x = a cos3 t, y = a sin3 t , then is
dx
(A) – tant (B) – cos t
(C) – cosec t (D) – cot t
dy dx
Solution: = 3a sin2 t cos t, = − 3a cos2 t sin t
dt dt
dy 3a sin2 t cos t
= = − tan t
dx − 3a cos 2 t sin t
 Hence (A) is correct answer.

dy
8. x = t2 + t + 1, y = t2 – t + 1; then is
dx
2t + 1 2t − 1
(A) (B)
2t − 1 2t + 1
2t + 1
(C) (D) none of these
(2t − 1)2
dx dy
Solution: x = t2 + t + 1, = 2t + 1, = 2t − 1
dt dt
dy 2t − 1
=
dx 2t + 1
Hence (B) is correct answer.

dy
9. x = a (1 – sint), y = a(1 – tan t) , then is
dx
(A) sec2t (B) sec3t
(C) cos2t (D) cos3t
dx
Solution: x = a (1 − sin t)  = a (0 − cos t) = − a cos t
dt
dy − a sec 2 t
= = sec3 t
dt − a cos t
Hence (B) is correct answer.

dy
10. y = loga(x2) , then at x = e, is
dx
(A) 2e (B) 1/e
(C) 2/e (D) none of these
2
dy log x 1 2x 2
Solution: y = loga x2 = = = 2 =
dx loge a x loge a x loge a
Hence (D) is correct answer.

dy
11. y = logsin −1 x cos −1 x , then is
dx
1 sin−1 x
(A) cos −1 x (B)
1 − x2 cos −1 x
tan−1 x
(C) log x (D) none of these
1 − x2
log cos −1 x
Solution: y = logsin −1 x cos −1 x, =
log sin −1 x
 dy 1 1 −1
= −1
 −1
 + log(cos−1x)  − 1 
dx log sin x cos x 1 − x2
1 1 1
−1 2
 −1

(log sin x ) sin x 1 − x2
hence none of these
Hence (D) is correct answer.

 eax − e −ax  dy
12. y = sin −1 ax − ax 
 , then is
e +e  dx
eax + e −ax 2a
(A) (B) ax
2a e + e − ax
4a eax + e − ax
(C) ax (D)
e + e − ax eax − e − ax
 eax − e −ax 
Solution: y = sin−1  ax ,
− ax 
e +e 
dy 1  a(eax + e −ax ) (eax − e −ax )  −1 a(eax − e −ax ) 
=   ax − ax
+ 
dx (eax − e − ax )2  (e + e ) (eax + e − ax )2 
1 − ax − ax 2
(e + a )
(eax + e − ax )  (eax + e −ax )2 − (eax − e −ax )2 
= a 
e 2ax + y − 2ax + 2 − e2ax − e − 2ax +2  (eax + e − ax )2 
a
( 2 + 2)
2a
= 2ax = ax
− ax
(e + e ) e + e − ax
Hence (B) is correct answer.

 y−x2 
tan − 1  
 x2


 dy
13. x= e , then is
dx
 y 2 2y  2y  y 2 2y 
(A) x 2 + 4 − 2  + (B) x 2 + 4 − 2 
 x x  x  x x 
 y 2 2y 
(C)  2 + 4 − 2  (D) none of these
 x x 
x = e tan (( y − x ) / x )
−1 2 2
Solution:
 y  y
log x = tan−1  2
− 1  2 − 1 = tan log x
x  x
y
= 1 + tan log x
x2
y = x2 + x2 tan log x
 dy 1
= 2x + 2x tan log x + x sec2 (log x) 
dx x
= 2x + 2x tan log x + x sec2 (log x)
2 2 2y
= (x + x2 tan log x) + x sec2 (log x) = + 1 + tan2 (log x)
x x
2y   y   2y
2
y2 2y
 + x 1 +  2 − 1  = + 1+ 4 + 1− 2
x  x   x x x

2
xy 2yx 2y
= 4 − 2 + + 2x
x x x
y 2 2y 2y y2
= 3− + + 2x = 3 + 2x
x x x x
Hence (A) is correct answer.

dy
14. x = (log x)tany , then is
dx
(A) logx – tany (B) logx + tany
log x − tan y log x − tan y
(C) (D)
x log x sec 2 y x log x sec 2 y. log log x
Solution: x = (log x)tany  log x = tan y log log x
log x
tan y = .
log log x
Hence (D) is correct answer.

n dy
15. If y = (1 + x) (1 + x 2 )(1 + x 4 ).......(1 + x 2 ) then at x = 0 is
dx
(A) 1 (B) 0
(C) –1 (D) none of these
n
Solution: y = (1 + x) (1 + x2 )(1 + x 4 ).......(1 + x2 )
Multiplying numerator and denominator by (1 – x)
(1 − x)(1 + x)(1 + x 2 ).........(1 + x 2n )
 y=
(1 − x)
 1− x 
n +1
2
 y=
 (1 − x) 
 

 =

n +1 2 −1
dy (1 − x). −2 .x
n +1
− 1− x2  ( n +1

) (−1)
dx (1 − x)2
dy −2n+1.0.1 + 1 − 0
So = =1
dx x =0 12
Hence (A) is correct answer.
 tanx + secx - 1
16. If f(x) = then f’(x) is equal to
tanx - secx + 1
(A) sec x(tanx – secx) (B) sec x(secx – tanx)
(C) sec x(tanx + secx) (D) none of these

 tan x + sec x − 1   tan x + sec x + 1 

Solution: f(x) =   
 tan x − sec x + 1   tan x + sec x + 1 
tan2 x + sec 2 x + 2 tan x sec x − 1
= = tan x + sec x
(1 + tan x)2 − sec 2 x
Thus f’(x) = sec2x + secx tanx = secx (secx + tanx)
Hence (C) is correct answer.

-1 1
17. The differential coefficient of cosec 2x 2 - 1 with respect to 1 − x 2 at x =
½ is
(A) –4 (B) 4
(C) –1 (D) none of these
 1  
Solution: Let x = cos; then y = cos−1  2  = cos−1(sec 2) = − 2
 2x − 1  2
z = 1 − x 2 = 1 − cos2  = sin 
dy dy d −2 −2
 = = =
dz dz d cos  x
Hence (A) is correct answer.

 5cosx - 12sinx   π dy
18. If y = cos -1   , x   0,  , then is equal to
 13   2 dx
(A) 1 (B) –1
(C) 0 (D) none of these

 5 12 
Solution: y = cos−1  cos x − sin x  = cos −1(cos  cos x − sin  sin x)
 13 13 
= cos ( cos(x + )) = x + 
−1

dy
 =1
dx
Hence (A) is correct answer.
dy
19. If sin(x + y) = ex+y − 2, then is equal to
dx
(A) x=0 (B) x=2
(C) x = −2 (D) x=3
Solution: If sin(x + y) = e x + y –2
then, x + y = constant = (say)
 where '' is the solution of sin = e – 2.
dy
Thus, =–1
dx
20. f(x) = (x2 − 4) x 2 − 5x + 6 + cos ( x ) is non-differentiable at
(A) x=0 (B) x=2 (C) x = −2 (D) x=3
Solution: 2
f(x) = (x – 4) |(x – 2) (x – 3)| + cosx
(x 2 − 4)(x 2 − 5x + 6) + cos x, x  2 or x  3
f(x) = 
 (4 − x )(x − 5x + 6), x  (2, 3)
2 2

only points where f(x) may be non–differentiable are x = 2 and x = 3.

(x 2 − 4)(2x − 5) + 2x(x 2 − 5x + 6) − sin x, x  2 or x  3
f(x) = 
 (4 − x )(2x − 5) − (x − 5x + 6)(2x) − sin x, x  (2, 3)
2 2

f(2 – 0) = – sin2,
f(2 + 0) = – sin2,
f(3 – 0) = –5 – sin3
f(3 + 0) = 5 – sin3
Thus, f(x) is differentiable at x = 2 but not at x = 3.

21. If f(x) = logx(logx), then f'(x) at x = e is equal to:

(A) e (B) 1/e
(C) 2/e (D) 0
log (log x ) logt
Solution : y = logx (log x ) = = ,
log x t
where t = log x
 when x = e, t = 1
1
t × - (logt )
dy 1 - logt
= t 2 - 1=
dt t t2
dy dy dt 1 − logt 1
 =  = 
dx dt dx t2 x
dy 1 - log1 1 1
= ´ = .
dx x =e (1)2 e e
Hence (B) is correct answer.

 sin x 2
 , x  1, then at x = 0
22. f(x) =  x
0, x = 0

(A) f(x) is continuous but non−differentiable
(B) f(x) is differentiable
 (C) f(x) is discontinuous at x = 0
(D) none of these

f(h) − f(0) sinh2

Solution: f(0) = lim = lim 2 = 1
h →0 h h →0 h
Thus f(x) is differentiable at x = 0 at hence also continuous at x = 0.

23. Exhaustive set of points where f(x) = x x is differentiable, is

(A) ( −, ) {0} (B) ( −, ) {0}
(C) ( −, )  −1 (D) ( −, )
x2, x0
Solution: f(x) =  2
− x , x  0
Graph of f(x) indicates that f(x) is differentiable for (–, ).
y
y = x2

O x

y =− x 2

 x
 2x 2 + x ,x  0
24. Let f(x) =  , then f(x) is
1 ,x=0

(A) Continuous but non−differentiable at x = 0
(B) Is differentiable at x = 0
(C) Discontinuous at (D) none of these

Solution: f(0 + 0) = lim f( −h)

h →0

h
= lim 2
h→0 2h + h

1
= lim =1
h→0 2h + 1

and f(0 – 0) = lim f( −h)

h →0

−h
= lim
h→0 2h + | −h |
2
 −h
= lim
h→0 2h2 + h

−1
= lim = –1
h→0 2h + 1

as f(0 + 0)  f(0 – 0), thus f(x) is discontinuous at x = 0.

FROM AIEEE BOOK

25. If f(x) is a polynomial in x, the second derivative of f(ex) at x = 1 is :
(A) ef"(e) + f'(e) (B) (f"(e) + f'(e))e2
(C) 2
e f"(e) (D) (f"(e)e + f'(e))e
Solution. x
If y = f(e ) (a polynomial in e ), then x

dy
= f ' (e x ).e x
dx
d2 y

dx 2 ( ) ( )
= e x f " e x .e x + e x f ' e x

d2 y

dx 2
( ) ( )
= e2x f " e x + e x f ' e x

d2 y
 2
= e2 f " ( e) + ef ' ( e)
dx x =1
Hence (D) is correct answer.

 sin x + cos x 
26. The differential coefficient of tan1−  w.r.t. x is :
 cos x − sin x 
1
(A) 0 (B)
2
(C) 1 (D) None of these
 sin x + cos x   1 + tan x 
Solution. y = tan−1   = tan−1 
 cos x − sin x   1 + tan x 
   
= tan−1  tan  + x  = + x
 4  4
dy
 =1
dx
Hence (C) is correct answer.

27. The differential coefficient of log (tan x) is :

(A) 2 sec 2x (B) 2 coesec 2x
(C) 2sec2x (D) 2cosec2x
d 1
Solution. log (tan x ) = .sec 2 x
dx tan x
 cos x 1 1
= . 2
=
sin x cos x sin x cos x
2 2
= = = 2cos ec 2x
2sin x cos x sin2x
Hence (B) is correct answer.

 1− x2  dy
28. If y = sin−1  2
, then is :
 1+ x  dx
-2 2
(A) (B)
1+ x2 1+ x2
1 2
(C) (D)
2 + x2 2 - x2
Solution. Pur x = tan 
  
 y = sin–1(cos 2) = sin−1  sin  − 2  
 2 
 
= − 2 = − 2 tan−1 x
2 2
dy −2
 =
dx 1 + x 2
Hence (A) is correct answer.

 1  1
29. Derivative of sec −1  2  w.r.t. 1 - x 2 at x = is :
 2x − 1 2
(A) 2 (B) 4
(C) 1 (D) –2
 1 
Solution. Let u = sec −1  2  , v = 1 − x 2
 2x − 1
Put x = cos  u = 2, v = sin
du dv du 2 2
 = 2, = cos   = =
d d dv cos  x
du
 = 4.
dv x = 1
2
Hence (B) is correct answer.

dy
30. If sin(x + y) = log(x + y), then =
dx
(A) 2 (B) –2
(C) 1 (D) –1
 dy  1  dy 
Solution. cos ( x + y)  1 +  = 1+
 dx  x + y  dx 
  1   dy 
 cos ( x + y ) − 1+ =0
 x + y   dx 
dy
 = −1
dx
Hence (D) is correct answer.

dy
31. If y = sin x + sin x + sin x + ....to , then value of is :
dx
sin x sin x
(A) (B)
y +1 y +1
cos x cos x
(C) (D)
2y + 1 2y - 1
Solution. y = sin x + y  y2 = sinx + y
dy dy dy
 2y = cos x +  ( 2y − 1) = cos x
dx dx dx
dy cos x
 =
dx 2y − 1
Hence (D) is correct answer.

d2 y
32. If y = log (sin x), then is :
dx 2
(A) –cosec2x (B) sec2x
(C) –cosecx cot x (D) secx tanx
dy cos x
Solution. y = log (sin x)  = = cot x
dx sin x
d2 y
 = − cos ec 2 x
dx 2
Hence (A) is correct answer.

 1− x2  dy
33. If y = log  2
, then is :
 1+ x  dx
- 4x 3 4
(A) (B)
1- x4 1- x4
- 4x 4x 3
(C) (D)
1- x4 1- x4
Solution. y = log (1 – x2) – log(1 + x2)
dy −2x 2x −4x
 = − =
dx 1 − x 1 + x
2 2
1− x4
Hence (A) is correct answer.
 dy
34. If y = a(1 + cos t) and x = a(t – sint), then is :
dx
t t
(A) tan (B) - tan
2 2
t
(C) - cot (D) None of these
2
dy
dy dt - a sin t
Solution. = =
dx dx a (1 - cos t )
dt
t t
- 2a sin cos
= 2 2 = - cot t
t 2
2a sin2
2
Hence (C) is correct answer.

x +.....to ¥ dy
35. If y = ex +e , then is :
dx
y y
(A) (B)
y +1 y- 1
y
(C) (D) None of these
1- y
Solution. y=e x+y  log y = x + y
1 dy dy dy y
 = 1+  =
y dx dx dx 1 − y
Hence (C) is correct answer.

dy  1 1
36. If x + y = 1, then at  ,  is :
dx  4 4
1
(A) (B) 1
2
(C) –1 (D) 2
1 1 dy
Solution. x + y =1  + .=0
2 x 2 y dx
dy x dy
 =−  = −1
dx y dx  1, 1 
 4 4

Hence (C) is correct answer.

d2 y
37. If x = at2, y = 2at, then is :
dx 2
 1 1
(A) - (B)
t2 2at 2
1 1
(C) - 3 (D) -
t 2at 3
dx
Solution. x = at2  = 2at
dt
dy
y = 2at  = 2a
dt
dy 2a 1 d2 y d  1 1 dt
\ = =  =   =− 2
dx 2at t dx 2
dx  t  t dx
 1  1 
= − 2  [By (1)]
 t   2at 
1
=-
2at 3
Hence (D) is correct answer.

1
38. Let g(x) be the inverse of the function f(x) and f'(x) = . Then g'(x) is :
1 + x3
1 1
(A) (B)
1 + (g (x )) 1 + (f (x ))
3 3

(C) 1 + (g(x))3 (D) 1 + (f(x))3

Solution : Since g(x) is the inverse of f(x)
 (gof)(x) = x
 g(f(x)) = x
 g'(f(x)) f'(x) = 1
1
Þ g' (f (x )) = = 1 + x3
f ' (x )
 g'(y) = 1 + (g(y))2
[ y = f(x)  x = f–1(y) = g(y)]
 g'(x) = 1 + (g(x))3. (Replacing y by x)
Hence (C) is correct answer.

 
d  2 −1 1 
39.  sin cot  =
dx  1+ x 
 
1− x 
1
(A) 0 (B) -
2
1
(C) (D) –1
2
Solution : Put x = cos    = cos–1 x
1 1
\ = sin2 cot - 1 sin2 cot - 1
1 + cos q q
cot
1 - cos q 2
   
= sin2 cot −1  tan  = sin2 cot −1 cot  − 
 2  2 2
  
= sin2  −  = cos2
 2 2 2
1 + cos q 1 + x
= =
2 2
dy 1
 = .
dx 2
Hence (C) is correct answer.

40. If f(x) = loga loga(x), then f'(x) is :

loga e loge a
(A) (B)
x loge x x loga x
loge a x
(C) (D)
x loge a
1 d
Solution : f ' (x ) = ×(loga e) × (laga x )
loga x dx
loga e 1 loga e loga e loga e
= × ×loga e = × = .
loga x x x loga x x loge x
Hence (A) is correct answer.

dy
41. If yx = xy, then is :
dx
y y (x log y - y )
(A) (B)
x x (y log x - x )
x x log y
(C) (D)
y y log x
Solution : y x = x y  log y x = log x y
 x log y = y log x
1 dy 1 dy
 x + log y  1 = y  + log x 
y dx x dx
x  dy y
  y − log x dx = x − log y
.
 dy y ( y − x log y )
Hence (B) is correct ans  = wer.
dx x ( x − y log x )

dp
42. If pv = 81, then at v = 9 is :
dx
(A) 1 (B) –1
(C) 2 (D) None of these
81 dp 81
Solution : pv = 81  p =  =− 2
v dv v
dp 81
 For v = 9, =− = −1.
dv 81
Hence (B) is correct answer.

 2x  2x
43. Differential coefficient of tan−1  2
w.r.t. sin- 1 is equal to :
 1− x  1+ x2
(A) 0 (B) –1
(C) 1 (D) None of these
 2x  2x
Solution : Put y = tan−1  2
and z = sin- 1
 1− x  1+ x2
Put x = tan    = tan−1 x
 y = tan−1 ( tan2 ) = 2, z = sin−1 ( sin2 ) = 2
dy dz dy
= 2, =2  = 1.
d d dz
Hence (C) is correct answer.

44. If f (x ) = 1 + x for x > 0, then f'(x) is :

1 1
(A) (B)
2f (x ) 4 xf (x )
2 x +1 1
(C) (D)
4 xf (x ) 2 xf (x )
 11 
Solution : f ' ( x) =  0 + 
2 1+ x 2 x
1 1
= = .
4 x 1+ x ( )
4 x ×f (x )

Hence (B) is correct answer.


45. If f(x) = cot–1 (cos2x)1/2, then f '   is :
 6
 1 2
(A) (B)
3 3
2 -2
(C) (D)
3 3
-1 1
f ' (x ) = × (cos 2x ) (- sin2x ×2)
- 1/ 2
Solution :
1 + cos 2x 2
sin2x
=
cos 2x (1 + cos 2x )

sin
 3
 f '  =
 6  
cos  1 + cos 
3  3
3
2 2
= = .
1 1 3
 1+ 
2  2
Hence (C) is correct answer.

 x + 1  x − 1 dy
46. If y = sec −1   + sin −1
  , then dx is :
 x − 1  x + 1
(A) 1 (B) 2
(C) 3 (D) 0
 x + 1  x − 1
Solution : y = sec −1   + sin −1
 
 x − 1  x + 1
 x − 1  x − 1 
= cos−1   + sin −1
 =
 x + 1  x + 1 2
dy
 =0.
dx
Hence (D) is correct answer.

dy
47. If x = a(t + sint), y = a(1 – cost), then is :
dx
t t
(A) tan (B) cot
2 2
t t
(C) sec (D) co sec
2 2
dx dy
Solution : = a (1 + cos t ), = a sin t
dt dt
 dy dy dt a sin t
 =  =
dx dx dx a (1 + cos t )
t t
2sin cos
= 2 2 = tan t .
t 2
2cos2
2
Hence (A) is correct answer.

48. If x2 + y2 = 1, then :
(A) yy" – (2y')2 + 1 = 0 (B) yy" + (y')2 + 1 = 0
(C) yy" – (y')2 – 1 = 0 (D) yy" + 2(y')2 + 1 = 0
Solution : y 2 = 1 − x 2  2y y ' = −2x
 y y ' = − x  y y ''+ y '+ y ' = −1
 y y ''+ ( y ') + 1 = 0 .
2

Hence (B) is correct answer.

LEVEL II

49. If f'(3) = 5 then lim

( ) (
f 3 + h2 − f 3 − h 2 ) is :
h →0 2h2
(A) 5 (B) 1/5
(C) 2 (D) None of these

Solution : lim
( ) (
f 3 + h2 − f 3 − h 2 )
2
h→0 2h

= lim
( )
f 3 + h2 − f ( 3) + f ( 3) − f 3 − h2 ( )
2
h→0 2h

=
1
lim
( 2
)
f 3 + h − f ( 3)
+
1
lim
(
f 3 − h2 − f ( 3 ) )
2 h→0 h2 2 h→ 0 −h2
1 1 1 1
= f ' (3) + f ' (3) = (5) + (5) = 5 .
2 2 2 2
Hence (A) is correct answer.
f ( a + h ) − 2f ( a ) + f ( a − h )
50. If f is twice differentiable function then lim is :
h→0 h2
(A) 2f'(a) (B) f''(a)
(C) f'(a) (D) f'(a) + f''(a)
f ( a + h) − 2f ( a) + f ( a − h)  0 
Solution : lim  0 
h→0 h2
f ' ( a + h) − f ' ( a − h)  0 
= lim  0 
h→0 2h
 f '' ( a + h) + f '' ( a − h)
= lim = f '' ( a) .
h→ 0 2
Hence (B) is correct answer.

51. If f(x) = sinx, g(x) = x2, h(x) = logx and F(x) = (hogof)(x) then F''(x) is :
(A) 2cosec3x (B) 2cotx2 – 4x2cosec2x2
(C) 2x cotx 2 (D) – 2cosec2x
Solution : (
F (x ) = (hogof )(x ) = h g (f (x )) )
= h (g (sin x )) = h (sin2 x ) = loge (sin2 x )
= 2 loge (sin x )
 F' ( x) 2cot x  f '' ( x) = −2 cosec 2 x
Hence (D) is correct answer.

52. ( )
If y = log x + 1 + x 2 then the value of y2(0) is :
(A) –1 (B) 0
(C) 1 (D) None of these

Solution : (
y = log x + x 2 + 1 )
1  1 
 y1 =  1 + ( 2x)
x + x2 + 1 2 x2 + 1 
1 −x
 y1 =  y2 =
( )
3/2
x2 + 1 1+ x2
 y2 ( 0) = 0 .
Hence (A) is correct answer.

( ) then (1+ x ) y
m
53. If y = x + 1 + x 2 2
2 + xy1 − m2 y =
(A) 0 (B) 1
(C) –1 (D) 2
m −1  1 2x 
Solution : y1 = m  x + 1 + x 2   1+ 
   2 1+ x2 
my
=
1+ x2
 ( )
y12 1 + x 2 = m2 y 2
Differentiating w.r.t. x,
2y1y 2 (1 + x 2 ) + y12 (2x ) = 2m2 yy1
Canceling 2y1,
 (1+ x ) y 2
2 + xy1 = m2 y.
Hence (A) is correct answer.

d2 y
54. If x + y + y − x = c then is :
dx 2
2 −2
(A) (B)
c c2
2
(C) (D) None of these
c2
Solution : We are given x+y + y- x =c . . . (1)

( ) -( )
2 2
Also x+y y- x = x + y - (y - x )

 ( x+y + y−x )( x + y − y − x = 2x )
2x
 x+y − y−x = [By (1)] . . . (2)
c
2x
(1) + (2) gives 2 x+y =c+
c
4x 2
Squaring 4 (x + y ) = c + 2 + 4x 2

c
2
4x
Canceling 4x, 4y = c 2 + 2
c
dy 8x d2 y 2
 4 = 2  2
= 2.
dx c dx c
Hence (C) is correct answer.

f '' ( 0 ) 1 n
If f ( x ) = (1 + x ) then the value of f ( 0 ) + f ' ( 0 ) + f ( 0 ) is :
n
55. + ....... +
2! n!
(A) n (B) 2n
(C) 2n−1 (D) None of these
f (0) = 1, f ' (x ) = n (1 + x ) ,
n- 1
Solution :
f '' (x ) = n (n - 1)(1 + x )
n- 2
, ......,
(n)
f(x) = n (n - 1) ......1 = n!
 f ' ( 0) = n, f '' ( 0) = n (n − 1) , ...., f n ( 0) = n!
n n (n - 1) n!
 Given expression = 1 + + + ...... +
1 2! n!
n n n n n
= C0 + C1 + C2 + ...... + Cn = 2 .
Hence (B) is correct answer.
 d2 y dy
56. If y = sin (sinx), and + tan x + f(x) = 0 then f(x) is :
dx 2 dx
(A) sin2x sin(cosx) (B) sin2x cos(sinx)
(C) 2
cos xsin(cosx) (D) cos2x sin(sinx)
dy
Solution : = cos (sin x ) ×cos x
dx
d2 y
 = − cos ( sin x )  sin x + cos x  − sin ( sin x ) cos x 
dx 2
d2 y dy
 + tan x = − cos ( sin x )  sin x − cos2 x  sin ( sin x ) + cos ( sin x )  cos x  tan x
dx 2 dx
= - cos2 x ×sin (sin x )
 f ( x) = cos2 x  sin ( sin x) .
Hence (D) is correct answer.
d2 y 
57. If x = 2 cos t – cos 2t and y = 2 sin t – sin 2t then the value of 2
at t =
dx 2
is :
(A) 3/2 (B) –5/2
(C) 5/2 (D) –3/2
dx
Solution : = - 2sin t + 2sin2t
dt
dy
= 2cos t - 2cos 2t
dt
dy 2cos t - 2cos 2t cos t - cos 2t
= =
dt - 2sin t + 2sin2t sin2t - sin t
3t t
2 sin sin
= 2 2 = tan 3t
3t t 2
2cos sin
2 2
2
dy 3t 3 dt
2
= sec 2 ´ ´
dx 2 2 dx
3 3t 1
= sec 2 ×
2 2 2sin2t - 2sin t
2
dy 3
 2
=− .
dx t =  2
2

Hence (D) is correct answer.

2
 dy 
58. If x = sec  − cos , y = sec  − cos  then   is equal to :
n n

 dx 
 (A)
(
n2 y 2 + 4 ) (B)
(
n2 y 2 − 4 )
x +42
x 2

2
y −4 2
 ny 
(C) n
x2 − 4
(D)  x  −4
 
dy
Solution : = nsec n−1   sec   tan  − n  cosn−1   ( − sin  )
d
 sin 
= n sec n  + cosn−1   sin 
 cos  
nsin 
= sec n  + cosn  
cos 
(
n tan  sec n  + cosn  )
dx sin 
= sec  tan  + sin  = sec  + sin 
d cos 
sin 
= ( sec  + cos ) = tan ( sec  + cos )
cos 


dy n tan sec  + cos 
=
n
( n
)
dx tan ( sec  + cos  )

=
(
n sec n  + cosn  )
sec  + cos 
( )
2
 dy 
2
n2 sec n  + cosn 
   =
 dx  ( sec  + cos  ) 2
Hence (A) is correct answer.

 2x − 1  dy
59. If y = f  2  and f'(x) = sin x2 then is:
 x + 1 dx
(A) cos x2.f'(x) (B) – cos x2.f'(x)

(C)
(
2 1+ x − x 2

sin
)
 2x − 1 
2

(D) None of these

 x2 + 1 
( )
2
x +1
2  
2x − 1
Solution : Let = z  y = f ( z)
x2 + 1
dy dz
\ = f ' (z ) ×
dx dx
dz
= sin z2 ×
dx
( f ' (z ) = sin z 2 )
2
 2x − 1 d  2x − 1
= sin  2
 x + 1 dx  x 2 + 1
 = sin  2
2
(
 2x − 1 2 1 + x − x
2

.
)
 x + 1 (
x2 + 1
2
)
Hence (C) is correct answer.
d  3 d2 y 
60. If y2 = p(x), a polynomial of degree 3 then 2 y  is equal to:
dx  dx 2 
(A) p'''(x) + p'(x) (B) p'''(x) + p''(x)
(C) p(x) p'''(x) (D) a constant
Solution : y = p ( x)  2y y1 = p' ( x)
2

 2 ( y y 2 + y1 y1 ) = p'' ( x )
1
 y y2 =
2
(
p'' ( x ) − 2y12 )
1
(
 y 3 y 2 = p'' ( x ) y 2 − 2 ( y y1 )
2
2
)
2
1 1 
= p'' ( x )  p ( x ) −  p' ( x )
2 2 
d 3 1 1  1 

dx
( 2
)
y y 2 = p''' ( x) p ( x) + p' ( x ) p'' ( x )  − 2   p' ( x )   p'' ( x )
2  2 
d 3
 2
dx
( )
y y 2 = p ( x ) p''' ( x ) .

=
((
n2 sec n  − cosn  )
2
+4 ) = n ( y + 4) .
2 2

( sec  − cos  ) 2 + 4 x2 + 4
Hence (C) is correct answer.