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STUDY PACKAGE
Subject : Mathematics
Topic: Differentiation
1.
2.
3.
4.
5.
6.
Index
Theor y
Question Bank
Exercise
Que. from Compt. Exams
8 Yrs. Que. from AIEEE
32 Yrs. Que. from IIT-JEE
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2 of 18
First Principle Of Differentiation
The derivative of a given function f at a point x = a on its domain is defined as:
Limit f (a + h)−f (a ) , provided the limit exists & is denoted by f ′(a).
h→0
h
f ( x )−f (a)
, provided the limit exists.
i.e. f ′(a) = Limit
x →a
x−a
2.
If x and x + h belong to the domain of a function f defined by y = f(x), then
Limit f ( x + h)−f ( x ) if it exists, is called the Derivative of f at x & is denoted by f ′(x) or dy . i.e., f ′(x) = Limit
h→0
h→0
dx
h
f ( x + h)−f ( x )
This method of differentiation is also called ab-initio method or first principle.
h
Solved Example # 1 Find derivative of following functions by first principle
(i)
f(x) = x2
(ii)
f(x) = tan x
(iii)
f(x) = esinx
2
2
2
( x + h) − x
lim 2xh + h = 2x.
Solution (i)
f′(x)
= hlim
=
→0
h→0
h
h
(ii)
(iii)
f′(x)
f′(x)
tan( x + h) − tan x
= hlim
→0
h
tan( x + h − x )[1 + tan x tan( x + h)]
lim
= h→0
h
sin ( x + h )
sin x
−e
e
= hlim
→0
h
e sin ( x + h )− sin x − 1  sin( x + h) − sin x 
sin x


= hlim
e
→0
h

sin( x + h) − sin x 
[
3.
4.
]
lim sin( x + h) − sin x
h
Differentiation of some elementary functions
f(x)
f′(x)
= esin x
tan h
= hlim
. (1 + tan2x) = sec2x.
→0
h
h→0
1.
2.
xn
ax
nxn – 1
ax An a
3.
An |x|
1
x
4.
logax
1
x An a
= esin x cos x
(x ∈ R, n ∈ R)
5.
sin x
cos x
6.
cos x
– sin x
7.
sec x
sec x tan x
8.
cosec x
– cosec x cot x
9.
tan x
sec2 x
10.
cot x
– cosec x
Basic Theorems
d
1.
(f ± g) = f′(x) ± g′(x)
dx
d
d
2.
(k f(x)) = k
f(x)
dx
dx
d
3.
(f(x) . g(x)) = f(x) g′(x) + g(x) f′(x)
dx
g( x ) f ′( x ) − f ( x ) g′( x )
d  f ( x ) 
4.
 g( x )  =
g2 ( x )
dx 

5.
d
(f(g(x))) = f′(g(x)) g′(x)
dx
This rule is also called the chain rule of differentiation and can be written as
Note that an important inference obtained from the chain rule is that
⇒
1
dy
=
dx / dy
dx
dy
dy dz
=
.
dx
dz dx
dy dx
dy
=1=
.
dx dy
dy
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1.
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A.
Differentiation
f(x)
=
=
(iii)
f(x)
=
f′(x)
=
1
2 sin (2x + 3)
x
.
d
(sin (2x + 3))
dx
1+ x2
(1 + x 2 ) − x(2x )
(iv)
=
(1 + x 2 )2
(vii)
Ans.
1+ x
1− x
sin x
1 + cos x
(i)6x (5x3 + x – 1)
(1 + x 2 )2
(v) cos4 x – 3 cos2 x sin2x
B.
cos3 x sin x
(viii)
An (sin x – cos x)
− x 2 + 2x + 1
2
( x − 2) ( x − 3 )
sin (2x + 3)
1− x2
(v)
(ii)
cos(2x + 3)
=
f(x) = x sin x
f′(x) = x. cos x + sin x
Solved Example # 3 If f(x) = sin (x + tanx) then find value of f′(0).
f′(0) = 2
Solution.
f′(x) = cos (x + tanx) (1 + sec2x)
Self Practice Problems :
1.
Find the derviative of following functions using first principle.
(i)
f(x) = x sin x
(ii)
f(x) = sin2 x
Ans. (i)
x cosx + sinx
(ii)
2sin x cos x
f
(
5
+ t ) − f (5 − t )
lim
2.
Evaluate if f′(5) = 7, then
Ans.
t→0
2t
3.
Differentiate the following functions
( x − 1)
(i)
(1 + 3x2) (2x3 – 1)
(ii)
( x − 2)( x − 3)
(iv)
3 of 18
sin ( 2x + 3 )
(iii)
2
(vi) ex ((sin x + cos x) x + sin x) (vii)
7.
(iii)
1+ x 2
(vi)
x ex sin x
x
2
1+ x
1
x
sec2
2
2
Derivative Of Inverse Trigonometric Functions.
y = sin–1 x
⇒
–
π
π
≤y≤
2
2
⇒
98930 58881
(ii)
(iv)
1
1/ 2
(1 + x )
(1 − x )3 / 2
(viii)
cos x + sin x
sin x − cos x
x = sin y
dx
= cos y
dy
dy
1
=
=
dx
cos y
1
1 − sin 2 y
dy
=
dx
1
1− x 2
– 1 < x < 1.
 π π
1 − sin 2 y but for values of y ∈  − ,  , cos y is always
 2 2
positive and hence the result. similarly let us find derivative of other inverse trigonometric functions.
Let
y = tan–1x
x = tan y
dx
dx
2
= sec2y = 1 + tan 2 y
⇒
dy = 1 + x
dy
dy
1
=
(x ∈ R)
dx
1+ x2
π
Also
if
y = sec–1x
y ∈ [0, π] –  
⇒
x = secy
2
dx
1
dy
dy
1
⇒
=
⇒
=
dy = sec y tan y
x. tan y
dx
dx
± x sec 2 y − 1
Note here that cos y ≠
1 − sin 2 y , rather cos y = ±
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another way of expressing the same concept is by considering y = f(x) and x = g(y) as inverse functions of each
other.
dx
1
dy
= f′(x)
and
= g′(y)
⇒
g′(y) =
dy
′
f
(
x)
dx
Solved Example # 2
Find the differential of the following functions with respect to x.
x
f(x) = sin( 2x + 3)
(iii)
f(x) =
(iv)
f(x) = x . sin x
(i) f(x) = esin x (ii)
1+ x2
sin
x
Solution.
(i)
f(x)
=e
d
f′(x)
= esin x
(sin x)
= esin x cos x
dx
C.
1.
Methods Of Differentiation
Logrithmic Differentiation
The process of taking logarithm of the function first and then differentiate is called Logarithmic Differentiation.
It is useful if
(i)
a function is the product or quotient of a number of functions OR
(ii)
a function is of the form [f(x)]g(x) where f & g are both derivable,
dy
Solved Example # 6
If y = xx find
dx
1
dy
dy
1
Solution.
An y = x An x ⇒
.
=
x
.
+
An
x
⇒
= xx (1 + An x)
y
dx
dx
x
dy
Solved Example # 7
If y = (sin x)An x, find
dx
Solution.
An y = An x . An (sin x)
1 dy
 An (sin x )

cos x
dy
1
+ cot x An x 
⇒
= (sin x)An x 
y dx = x An (sin x) + An x. sin x
x
dx


x1/ 2 (1 − 2x )2 / 3
dy
find
dx
( 2 − 3 x )3 / 4 (3 − 4 x ) 4 / 5
2
4
1
3
Solution.
An y =
An x +
An (1 – 2x) –
An (2 – 3x) –
An (3 – 4x)
3
5
2
4
1 dy
9
16
4
1
⇒
=
–
+
+
y dx
4 (2 − 3 x )
5 (3 − 4 x )
3(1 − 2x )
2x
 1

4
9
16
dy
= y  2x − 3 (1 − 2 x ) + 4(2 − 3 x ) + 5 (3 − 4 x ) 
dx


2.
Implicit differentiation
If f(x, y) = 0, is an implicit function then in order to find dy/dx, we differentiate each term w.r.t. x regarding y as a
functions of x & then collect terms in dy/dx.
dy
Solved Example # 9
If x3 + y3 = 3xy find
dx
Solution.
Differentiation both sides w.r.t.x, we get
y − x2
dy
dy
dy
2
2
3x + 3y
= 3x
+ 3y
= 2
y −x
dx
dx
dx
2
Note that above result holds only for points where y – x ≠ 0
Solved Example # 8
If
y=
4 of 18
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1

sec y > 1

1
dy
dy
 x x2 − 1
= 
⇒
=
x ∈ (– ∞, – 1) ∪ (1, ∞)
1
dx
dx
| x | x2 − 1
−
sec y < −1
 x x 2 − 1
results for the derivative of inverse trigonometric functions can be summarized as :
f(x)
f′(x)
1
;
|x| < 1
sin–1x
1− x 2
−1
cos–1x
;
|x| < 1
1− x 2
1
tan–1x
;
x∈R
1+ x2
−1
cot–1x
;
x∈R
1+ x2
1
sec–1 x
; |x| > 1
| x | x2 − 1
−1
cosec-1 x
; |x| > 1
| x | x2 − 1
Solved Example # 4
If f(x) = An (sin –1 x2) find f′(x)
2x
1
1
Solution.
f′(x)
=
.
. 2x
=
−
1
−1 2
(sin x 2 ) 1 − x 4
(sin x )
1 − ( x 2 –1
)2
–1
Solved Example # 5
If f(x) = 2x sec x – cosec (x) then find f′(–2)
2x
1
Solution.
f′(x) = 2 sec –1(x) –
+
2
| x | x −1
| x | x2 − 1
2
1
5
4π
f′(–2) = 2.sec–1(– 2) +
+
f′(–2) =
+
.
3
3
2 3
2 3
.........(i)
dy
dy
y
+ lnx
=1–
dx
dx
x
⇒
y
1−
dy
x
=
+
A
1
n
x
dx
⇒
x−y
dy
= x(1 + An x )
dx
dy
dx
du
dv
Solution.
u+v=2
⇒
+
=0
dx
dx
y
where u = x
&
v = yx
⇒
An u = y An x
&
An v = x An y
x dy
1 du
dy
y
1 dv
=
+ An x
&
= An y + y
⇒
u dx
dx
dx
x
v dx

x dy 
dy 
y
du
dv
 &
⇒
= x y  + An x
= yx  An y + y dx 
dx 
dx
dx
x


⇒

x dy 
dy 
y

 + y x  An y +
xy  + An x
y dx  = 0.
dx 
x

 x
y y
 y An y + x . 
dy
x

=–
dx
 y
x
 x An x + y x . 
y

⇒
Self Practice Problems
1.
Differentiate the following functions :
(i)
98930 58881
Solved Example # 11 If xy + yx = 2 then find
y = sec–1 (x 2)
(ii)
 1+ x 

y = tan–1 
 1− x 
(v)
y = (ln x)x + (x)sin x
x
1

y = 1 + 
x

dy
Find
if
dx
(i)
y = cos (x + y)
(iii)
2.
3.
4.
(ii)
x
x2/3 + y2/3 = a2/3
(iii)
x = y An (x – y)
An x
dy
=
.
dx
(1 + An x )2
x
a
x
dy
= log
, prove that
=2– .
If
x−y
x−y
y
dx
If xy = e x – y, then prove that
Ans.
1. (i)
2
x x4 − 1
(ii)
1
1+ x2
x
(iv) xx. e x (Anx + 1)
3.
y = ex
(iv)
x
1
1 
1  

(iii) 1 +  An 1 + x  + 1 + x 


x  


 1 

 sin x
+ cos x Anx 
(v)  An (Anx ) +  Anx   (An x) x + xsinx 



 x

1/ 3
− sin( x + y )
y( x − y )
y


2.
(i) 1 + sin( x + y )
(ii) –
(iii) x( x + y )
x
Differentiation using substitution
Following substitutions are normally used to sumplify these expression.
(i)
x 2 + a2
⇒
x = a tan θ
or
a cot θ
(ii)
a2 − x 2
⇒
x = a sin θ
or
a cos θ
(iii)
x 2 − a2
⇒
x = a sec θ
or
a cosec θ
(iv)
x+a
a−x
⇒
x = a cos θ
Solved Example # 12 :
Solution.
Let
 1 + x 2 − 1


Differentiate y = tan–1 
.
x


x = tan θ
 | sec θ | −1 

y = tan–1 
 tan θ 
⇒
θ = tan–1x
;
 π π
θ ∈ − , 
 2 2
 π π
[ |secθ | = secθ ∀ θ ∈  − ,  ]
 2 2
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Solution.
Taking log on both sides
y An x = (x – y)
differentiating w.r.t x, we get
dy
dx
5 of 18
If xy = ex – y, then find
Solved Example # 10
y=
θ
2
⇒
y=
1
tan–1 x
2
Solved Example # 13 :
Solution.
x = cosθ
θ = cos–1 (x)
θ ∈ [0, π]
;
y = tan–1
 1 + cos θ − 1 − cos θ 


 1 + cos θ + 1 − cos θ 


⇒
y = tan–1

 1 − tan


 1 + tan

⇒
y=
⇒
Note that
Also
π
1
–
cos–1x
4
2
2 cos
tan–1 (tan x) = x
Solution.
⇒
θ
2
θ
θ

 2 cos − 2 sin 
2
2

θ
θ

 2 cos + 2 sin 
2
2

⇒
y = tan–1
⇒
y=
⇒
1
dy
=
dx
2 1− x2
 π
θ
∈  0,  ,
 2
2
 π π
x ∈ − ,  .
 2 2
π
θ
–
4
2
2 cos
but for
for




θ
2
=
2 cos
θ
2
 2x 
If f(x) = sin–1 
 then find
 1+ x2 
 1
(ii)
f′  
(iii)
f′ (1)
2
Solved Example # 14
(i)
θ

2
θ

2
1 + cos θ =
6 of 18
⇒
⇒
f′ (2)
x = tan θ
θ = tan –1(x)
;
–
π
π
<θ<
2
2
⇒
y = sin–1 (sin 2θ)
2

−
x >1
π


< 2θ < π
1+ x 2
−1
 π − 2θ


π
−
>
2
tan
x
x
1
2
 2


−1< x < 1
−1
π
π

−
≤
≤
2
tan
x
1
x
1

− ≤ 2θ ≤
1+ x2
y =  2θ
⇒
f(x)
=
⇒
f′
(x)
=

2
2
− ( π + 2 tan −1 x )

x < −1
 −2
x < −1

− ( π + 2θ) − π < 2θ < − π
 1 + x 2

2
 1
2
8
(i)
f′(2) = –
(ii)
f′   =
(iii)
f′(1+) = – 1
&
f′(1– ) = + 1
2
5
5
 
⇒
f′(1) does not exist.
Aliter Above problem can also be solved without any substution also, but in a little tedious way.
 2x 

f(x) = sin–1 
 1+ x2 
1
f′(x) =
1−
=
f′(x) =
Solution.
⇒
(1 + x 2 )2
(1 + x 2 )2
(1 + x 2 )
2 2
(1 − x )
.
2(1 − x 2 )
(1 + x 2 )2
2
2
(1 + x 2 )
Solved Example # 15 If
.
4x 2
2{(1 + x 2 ) − 2x 2 }
.
(1 − x )
| 1− x2 |
1− x 2 +
thus
 2

1+ x2

f′(x) =  − 2
1 + x 2
|x|<1
|x|>1
1− y 2 = a(x – y), then prove that
Put
x = sin α
⇒
α=
y = sin β
⇒
β = sin–1 (y)
cosα + cos β = a (sinα – sinβ)
sin –1
(x)
dy
=
dx
1− y2
1− x2
.
98930 58881
 1 − cos θ 

y = tan–1 
 sin θ 
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θ

y = tan–1  tan 
2

 π π
[tan–1 (tanx) = x for x ∈  − ,  ]
 2 2
1
dy
⇒
=
2(1 + x 2 )
dx
 1+ x − 1− x
dy
Find
where y = tan–1 
dx
 1+ x + 1− x
⇒
1− y2
1− x 2
Aliter Using implicit differention.
y
−x
2 –
1− y2
1− x

y
 −
 a
1− y2

⇒
dy
=
dx
⇒
dy
=0
dx
dy
=
dx
4.
1− x2
 dy 
dy

= a 1 −
dx
 dx 

 dy
=a+

 dx
x
1− x
dy
=
dx
⇒
2
a+
a−
x
1− x2
y
1− y2
1− x 2 + 1− y 2
x
+
x−y
1− x 2
1− x2 + 1− y2
y
−
x−y
1− y2
(1 − x 2 ) + (1 − x 2 )(1 − y 2 ) + x 2 − xy
dy
=
.
dx
(1 − x 2 )(1 − y 2 ) + (1 − y 2 ) − xy + y 2
⇒
1− y2
dy
=
dx
⇒
98930 58881
1
–
α −β
 =a
cot 
 2 
⇒
1− y2
1− y2
1− x2
=
1 + (1 − x 2 )(1 − y 2 ) − xy
1 + (1 − x 2 )(1 − y 2 ) − xy
.
Solved Example # 16 If x= a cos3 t and y = a sin3t. Find
dy
dy
dx
Solved Example # 17 If y = a cos t and x = a (t – sint) find the value of
−a sin t )
dy
= a(1 − cos t )
dx
dy
dx t = π = – 1.
2
Derivative of one function with respect to another
dy dy / dx
f '(x)
=
=
Let y = f(x); z = g(x) then
.
d z d z / dx
g' (x)
Solved Example # 18
Find derivative of y = An x with respect to z = ex.
1
dy
dy / dx
=
=
dz
dz / dx
xex
Self Practice Problems :
dy
1.
Find
when
dx
(i)
x = a (cos t + t sin t) &
y = a (sin t – t cos t)
 1− t 2 
 2t 



(ii)
x=a 
&
y=b. 
2 
 1+ t 2 
 1+ t 
2.
3.
(i)
tan t
(ii)
dy / dθ
θ is a parameter, then dx = dx / dθ .
− 3a sin 2 t cos t
dy
dy / dt
=
=
= – tan t
3a cos 2 t sin t
dx
dx / dt
Ans.
1− x2
Hence proved
1− x2
Parametric Differentiation If y = f(θ) & x = g(θ ) where
5.
1− y2
( t 2 − 1)b
2at


x2
2 xa 2
dy


If y = sin–1  4
then
prove
that
=
.
4 
dx
x 4 + a4
 x +a 
 2x 
2
dy
 then prove that
If y = tan–1 
=
(| x | ≠ 1)
2
dx
 1− x 
1+ x2
dy
π
at t = .
dx
2
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1
α +β
α −β
 sin 


 2 
 2 
x – sin–1 y = 2 cot–1(a)
7 of 18
α +β
α −β
 cos 
 = 2a cos
2cos 
 2 
 2 
⇒
α – β = 2 cot–1 (a)
⇒
sin–1
differentiating w.r.t to x.
⇒
Derivatives of Higher Order
1 − u2
1− v 2
8 of 18
D.
du
=
dv
.
Let a function y = f(x) be defined on an open interval (a, b). It’s derivative, if it exists on (a, b) is a certain function
f ′(x) [or (dy/dx) or y′ ] & is called the first derivative of y w. r. t. x.
If it happens that the first derivative has a derivative on (a, b) then this derivative is called the second derivative of
y w. r. t. x & is denoted by f ′′(x) or (d2y/dx2) or y ′′ .
d3 y d  d2 y 
rd
Similarly, the 3 order derivative of y w. r. t. x, if it exists, is defined by 3 = dx  2  It is also denoted by f ′′′(x)
dx
 dx 
or y ′′′.
Solved Example # 19
If y = x3 An x then y′′and y′′′
1
Solution.
y′ = 3x2 An x + x3
x
y′ = 3x2 An x + x 2
1
y′′ = 6x An x + 3x2 .
+ 2x
x
y′′ = 6x An x + 5x
y′′′ = 6 An x + 11
x
Solved Example # 20
Solution.
 1
If y =   then find y′′(1)
x
An y = – x An x
when x = 1 ⇒
y=1
y′
⇒
⇒
y′ = – y (1 + An x)
y = – (1 + An x)
again diff. w.r.t. to x,
1
y
⇒
y′′ = y (1 + ln x)2 –
(using (i))
⇒
y′′ = – y′(1 + An x) – y .
x
x
It must be carefully noted that in case of parametric functions
dy
dy / dt
d
d2 y
d2 y / dt 2
d2 y
≠
although
=
but
rather
=
2
2
2
dx
dx / dt
dx
dx
dx / dt
dx 2
which on applying chain rule can be resolved as
 dx d2 y dy d2 x 
 .

−
 dt dt 2 dt . dt 2 
d2 y
d2 y
d  dy / dt  dt



.
⇒
.
2
2 = dt
2 =
dx
/
dt
dx


dx
dx
 dx 
 
 dt 
2
2
 dx d y dy d x 
.
 . 2 −

2
dt dt 2 
 dt dt
d y
=
3
dx 2
 dx 
 
 dt 
Solved Example # 21
Solution.
⇒
If x = t + 1 and y = t2 + t3 then find
dy
= 2t + 3t2 ;
dt
dy
= 2t + 3t2
dx
d2 y
dx 2
......(i)
y′′(1) = 0
 dy / dt 


 dx / dt 
dt
dx
.
dx
=1
dt
⇒
d2 y
dx
2
=
d
dt
(2t + 3t2) .
dt
dx
2
d y
dx 2
= 2 + 6t.
Solved Example # 22 If x = 2 cos t – cos 2t and y = 2 sin t – sin 2t then find value of
dy
dx
Solution.
= 2 cos t – 2 cos 2t
= 2 sin
dt
dt
3t
t
2 sin . sin
dy
cos t − cos 2t
2
2
=
=
.
3t
t
dx
sin 2t − sin t
2 cos . sin
2
2
d2 y
dy
d
3t
⇒
= tan
⇒
2 = dx
dx
2
dx
d2 y
dx
2
at t =
π
.
2
2t – 2 sin t
3t 

 tan 
2

⇒
d2 y
dx
2
=
98930 58881
If u = sin (m cos–1x) and v = cos (m sin –1 x) then prove that
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4.
d 
3 t  dt
 tan  .
dt 
2  dx
⇒
.........(ii) which can also be remembered as
d2 y
d x
dx 2
2 = –
3 .
dy
 dy 
 
 dx 
Solved Example # 25
y = sin (sinx) then prove that
2
y′′ + (tanx) y′ + y cos2x = 0
Solution.
Such expression can be easily proved using implict differention.
⇒
y′ = cos (sin x) cos x ⇒
sec x.y′ = cos (sin x)
again differentiating w.r.t x, we can get
secx y′′ + y′ sec x tan x = – sin (sin x) cos x
⇒
tanx y′ = – y . cos 2 x ⇒
y′′ +(tanx) y′ + y cos2x = 0
Self Practice Problems :
d2 y
2An x − 3
An x
1.
If y =
then find
Ans.
2
x
dx
x3
2.
Prove that y = x + tan x satisfies the differentiation equation
d2 y
cos2 x
– 2y + 2x = 0.
dx 2
sec 3 θ
d2 y
3.
If x = a (cosθ + θ sin θ) and y = a(sin θ – θ cosθ) then find
.
Ans.
aθ
dx 2
x sin x − cos x
4.
Find second derivative of Anx with respect to sin x.
Ans.
x 2 cos3 x
5.
if y = e– x (A cos x + B sin x), prove that
dy
d2 y
2 + 2 . dx + 2y = 0.
dx
If y = (tan–1x) 2 then prove that (1 + x2)2
Solved Example # 26
Solution.
⇒
⇒
dy
2 tan −1 x
=
dx
1+ x2
dy
(1 + x2)
= 2tan–1 (x)
dx
(1 + x2)
d2 y
dx 2
⇒
dy
+ 2x (1 + x2)
=2
dx
(1 + x2 )
d2 y
dx
2
d2 y
dx 2
+ 2x
+ 2x (1 + x2)
2
dy
=
(1 + x 2 )
dx
dy
= 2.
dx
9 of 18
98930 58881
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d  1  dx

 .
dx  f ′( x )  dy
f ′′( x )
=–
. g′(y)
f ′( x )2
f ′′( x )
g′′(y) = –
f ′( x )3
=
TEKO CLASSES, MATHS BY SUHAAG SIR PH: (0755)- 32 00 000,
3
3t
. sec 2
d2 y
3
2
2
⇒
=–
2
2 =
π
dx
−
2
(sin
2
t
sin
t
)
2
dx
t=
Solved Example # 23
Find second order derivative of 2y= sin x with respect to z = ex.
cos x
dy
dy / dx
=
=
Solution.
⇒
dz
dz / dx
ex
d2 y
d2 y
d  cos x 
d  cos x  dx
.
 x 

⇒
=
⇒
=
2
2
dz  e 
dx  e x  dz
dz
dx
1
− e x sin x − cos xe x
=
. x
x 2
e
(e )
2
(sin x + cos x )
d y
2 = –
e2x
dz
Solved Example # 24: y = f(x) and x =g(y) are inverse functions of each other than express g′(y) and g′′(y) in terms of
derivative of f(x).
dx
dy
Solution.
= f′(x) and
dy = g′(y)
dx
1
⇒
g′(y) =
...........(i)
again differentiating w.r.t. to y
f ′( x )
d  1 

g′′(y) = dy 
 f ′( x ) 
d2 y
10 of 18
11.
f ( x ) g( x ) h( x )
l( x ) m( x ) n( x )
If F(x) =
, where f, g, h, l, m, n, u, v, w are differentiable functions of x then F ′ (x)
u( x ) v( x ) w( x )
12.
L’ Hospital’s Rule:
If f(x) & g(x) are functions of x such that:
Limit f(x) = 0 = Limit g(x) OR Limit f(x) =
x →a
x →a
x →a
∞ = Limit
x →a g(x) &
(ii)
Both f(x) & g(x) are continuous at x = a &
(iii)
Both f(x) & g(x) are differentiable at x = a &
Both f ′(x) & g ′(x) are continuous at x = a, Then
f (x)
f ' (x)
f " (x)
Limit
Limit
Limit
x →a g( x ) = x →a g' ( x ) = x →a g" ( x ) & so on till indeterminant form vanishes
________________________________________________________________________________________________
(iv)
QUESTION BANK ON METHOD OF
DIFFERENTIATION
Select the correct alternative : (Only one is correct)
Q.1
If g is the inverse of f & f ′ (x) =
(A) 1 + [g(x)]5
Q.2
If
y = tan−1
(A) 2
Q.3
(B)
1
then g ′ (x) =
1+ x 5
1
1 + [g(x)]5
(C) −
1
1 + [g(x)]5
(D) none
 An e2 
d 2y
x 
−1 3 + 2 An x then

+
tan
=
 An ex 2 
dx 2
1 − 6 An x


(B) 1
(D) − 1
(C) 0
dy
 3x + 4 
=
 & f ′ (x) = tan x2 then
dx
 5x + 6 
If y = f 
2
(A) tan x3
Q.4
If y = sin−1  x 1 − x +
(B) sin−1 x
(A) 0
Q.5
(A)
Q.6
Q.8
(C) sin−1 x
(D) none
(D) none of these
dy
 2x − 1
If y = f  2  & f ′ (x) = sin x then
=
dx
 x + 1
1 + x − x2
(1 + x )
2
2
 2x − 1
sin  2

 x + 1
(B)
) sin  2x − 1 (C) 1 − x + x
(
 x + 1
(1 + x )
(1 + x )
2 1 + x − x2
2
2
Let g is the inverse function of f & f ′ (x) =
1 + a2
a 10
dy
If sin (xy) + cos (xy) = 0 then
=
dx
y
y
(A)
(B) −
x
x
dy 
2x
If y = sin−1
is :
2 then dx 
1+ x
x = −2
(A)
Q.7
 3 tan x 2 + 4 
 3x + 4 
1

 tan x2
.
(C)
f

2
2
5
tan
x
6
+
5
x
+
6



 (5x + 6)
dy
1
=
+ p, then p =
x 1 − x 2  &
dx
2 x (1 − x)
(B) − 2 tan 
5
210
(B)
2 2
2
x10
(1 + x )
2
(C)
2
. If g(2) = a then g ′ (2) is equal to
a 10
1 + a2
(C) −
 2x − 1
 (D) none
 x 2 + 1
sin 
x
y
(D)
1 + a 10
a2
x
(D) y
98930 58881
(i)
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f ' ( x ) g' ( x ) h' ( x )
f ( x ) g( x ) h( x )
f ( x ) g( x ) h( 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 )
Q.9
2
(B)
(A) 4
(B) 1/4
11 of 18
2
5
2
(C) −
5
5
1
 1 
The derivative of sec−1  2  w.r.t. 1 − x 2 at x = is :
2
 2x − 1
(A)
(D) none
(C) 1
(D) none
 d   3 d y
  y . 2  equals :
 dx  
dx 
If y2 = P(x), is a polynomial of degree 3, then 2 
Q.11
(A) P ′′′ (x) + P ′ (x) (B) P ′′ (x) . P ′′′ (x) (C) P (x) . P ′′′ (x)
(D) a constant
Let f(x) be a quadratic expression which is positive for all real x . If g(x) = f(x) + f ′ (x) + f ′′ (x), then for any real
x, which one is correct .
(A) g(x) < 0
(B) g(x) > 0
(C) g(x) = 0
(D) g(x) ≥ 0
Q.12
Q.13
dy
If xp . yq = (x + y)p + q then
is :
dx
(A) independent of p but dependent on q
(C) dependent on both p & q
g (x) . cos x1
0
if x ≠ 0
where g(x) is an even function differentiable at x = 0, passing through the
if x = 0
Let f(x) = 
origin . Then f ′ (0) :(A) is equal to 1
Q.14
If y =
1
1+ x
(A) emnp
Q.15
n− m
+x
p− m
Q.16
Q.17
1
1+ x
m− n
(B) 2
+x
p− n
+
(C) is equal to 2
1
1+ x
m− p
+x
n− p
then
(C) enp/m
(C) 4
If f is differentiable in (0, 6) & f ′ (4) = 5 then Limit
x→2
dy
at e m
dx
np
(D) does not exist
is equal to:
(D) none
c h=
f ( 4) − f x
(D) none of these
2
2−x
(A) 5
(B) 5/4
(C) 10
(D) 20
Lim+ xm (ln x)n where m, n ∈ N then :
Let l = x→
0
(A) l is independent of m and n
(B) l is independent of m and depends on m
(C) l is independent of n and dependent on m (D) l is dependent on both m and n
x
Let f(x) = 2 sin x x
2
1
f ′ (x)
=
2x . Then Limit
x→0
x
1
tan x
x
cos x
sin x
(A) 2
Q.19
(B) is equal to 0
log sin 2 x cos x
Lim
x has the value equal to
x→0
log 2 x cos
sin
2
2
cos x
Q.18
+
(B) emn/p
(A) 1
(B) dependent on p but independent of q
(D) independent of p & q both .
(B) − 2
cos x
(C) − 1
(D) 1
 π
Let f(x) = cos 2x sin 2x 2 cos 2x then f ′   =
2
cos 3x sin 3x 3 cos 3x
Q.20
(A) 0
(B) – 12
(C) 4
(D) 12
People living at Mars, instead of the usual definition of derivative D f(x), define a new kind of derivative, D*f(x) by
the formula
f 2 (x + h) − f 2 (x)
where f2 (x) means [f(x)]2. If f(x) = x lnx then
D*f(x) = Limit
h
h→ 0
D * f (x ) x = e has the value
(A) e
(B) 2e
(C) 4e
(D) none
Q.21 If f(4) = g(4) = 2 ; f ′ (4) = 9 ; g ′ (4) = 6 then Limit
x→4
(A) 3 2
Q.22
(B)
3
2
(C) 0
f (x) − g (x)
x −2
is equal to :
(D) none
f (x + 3h) − f (x − 2h)
If f(x) is a differentiable function of x then Limit
=
h→0
h
(B) 5f ′ (x)
(C) 0
(D) none
(A) f ′ (x)
98930 58881
Q.10
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2
d 2x
is :
dy 2
(B) −
(A) ex
(1+ e )
x
(C) −
3
ex
(1+ e )
x
(D)
2
d 2y
at the point (1, 1) is :
dx 2
3
5
(B) −
(C) −
8
12
−1
(1+ e )
x 3
If x2y + y3 = 2 then the value of
(A) −
3
4
(D) none
g (x) . f (a ) − g (a ) . f (x)
If f(a) = 2, f ′ (a) = 1, g(a) = − 1, g ′ (a) = 2 then the value of Limit
x→a
x−a
(B) 1/5
(C) 5
(D) none
(A) − 5
Q.26 If f is twice differentiable such that f ′′ (x) = − f (x), f ′ (x) = g(x)
Q.25
is:
98930 58881
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Q.24
ex
12 of 18
If y = x + ex then
h ′ (x) = [f (x)] + [g(x)] and
2
then the equation y = h(x) represents :
(A) a curve of degree 2
(C) a straight line with slope 2
Q.27
Q.28
Q.29
Q.30
Q.31
Q.32
Q.33
Q.34
h (0) = 2 , h (1) = 4
The derivative of the function, f(x)=cos-1
1 + x 2 at x =
2
3
is :
4
RS
T
(B) a curve passing through the origin
(D) a straight line with y intercept equal to − 2.
1
1
(2 cos x − 3 sin x) + sin−1
(2 cos x + 3 sin x)
13
13
UV
W
RS
T
UV w.r.t.
W
3
5
10
(A)
(B)
(C)
(D) 0
2
2
3
x
Let f(x) be a polynomial in x . Then the second derivative of f(e ), is :
(A) f ′′ (ex) . ex + f ′ (ex)
(B) f ′′ (ex) . e2x + f ′ (ex) . e2x
x
2x
(C) f ′′ (e ) e
(D) f ′′ (ex) . e2x + f ′ (ex) . ex
1
The solution set of f ′ (x) > g ′ (x), where f(x) = (52x + 1) & g(x) = 5x + 4x (ln 5) is :
2
(A) x > 1
(B) 0 < x < 1
(C) x ≤ 0
(D) x > 0
dy
x2 − 1
x2 + 1
−1
If
+ sec
, x > 1 then
is equal to :
2
2
dx
x +1
x −1
x2
x
(A) 4
(B) 4
(C) 0
x −1
x −1
dy
x x x x x x
If y =
...... ∞ then
=
dx
a + b+ a + b+ a + b+
a
b
a
(B)
(C)
(A)
ab + 2 by
ab + 2 by
ab + 2 ay
y = sin−1
(D) 1
(D)
b
ab + 2 ay
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.
y
If y = sin mx then the value of y 3
y6
y1
y4
y7
y2
y 5 (where subscripts of y shows the order of derivatiive) is:
y8
(A) independent of x but dependent on m
(C) dependent on both m & x
y ′′
If x2 + y2 = R2 (R > 0) then k =
(1 + y′ )
2 3
(B) dependent of x but independent of m
(D) independent of m & x .
where k in terms of R alone is equal to
1
2
1
2
(B) –
(C)
(D) – 2
2
R
R
R
R
Q.35 If f & g are differentiable functions such that g ′ (a) = 2 & g(a) = b and if fog is an identity function then f ′ (b)
has the value equal to :
(A) 2/3
(B) 1
(C) 0
(D) 1/2
(A) –
Q.36
x3
Given f(x) = −
+ x2 sin 1.5 a − x sin a . sin 2a − 5 arc sin (a2 − 8a + 17) then :
3
(A) f(x) is not defined at x = sin 8
(B) f ′ (sin 8) > 0
TEKO CLASSES, MATHS BY SUHAAG SIR PH: (0755)- 32 00 000,
Q.23
Q.39
Q.40
Q.41
Q.42
Q.43
(
Given : f(x) = 4x3 − 6x2 cos 2a + 3x sin 2a . sin 6a + An 2 a − a 2
(A) f(x) is not defined at x = 1/2
(B) f ′ (1/2) < 0
(C) f ′ (x) is not defined at x = 1/2
(D) f ′ (1/2) > 0
If
y = (A + Bx) emx + (m − 1)−2 ex
)
d 2y
dy
then 2 − 2m
+ m2y is equal to :
dx
dx
(B) emx
(C) e−mx
(D) e(1 − m) x
(A) ex
ax
bx
Suppose f (x) = e + e , where a ≠ b, and that f '' (x) – 2 f ' (x) – 15 f (x) = 0 for all x. Then the product ab is
equal to
(A) 25
(B) 9
(C) – 15
(D) – 9
Let h (x) be differentiable for all x and let f (x) = (kx + ex) h(x) where k is some constant. If h (0) = 5, h ' (0) = –
2 and f ' (0) = 18 then the value of k is equal to
(A) 5
(B) 4
(C) 3
(D) 2.2
Let ef(x) = ln x . If g(x) is the inverse function of f(x) then g ′ (x) equals to :
(A) ex
(B) ex + x
(C) e ( x + e x )
(D) e(x + ln x)
dy
The equation y2e xy = 9e–3·x 2 defines y as a differentiable function of x. The value of
for
dx
x = – 1 and y = 3 is
15
9
(B) –
(C) 3
(D) 15
(A) –
2
5
( )
(x x )
x
then :
Let f(x) = x x and g(x) = x
(A) f ′ (1) = 1 and g ′ (1) = 2
(B) f ′ (1) = 2 and g ′ (1) = 1
(C) f ′ (1) = 1 and g ′ (1) = 0
(D) f ′ (1) = 1 and g ′ (1) = 1
Q.45 The function f(x) = ex + x, being differentiable and one to one, has a differentiable inverse f–1(x). The value of
d –1
(f ) at the point f(l n2) is
dx
1
1
1
(A)
(B)
(C)
(D) none
An2
3
4
π
log sin|x| cos3 x
Q.46 If f (x) =
for |x| <
x≠0
3
3 x 
log sin|3x| cos  
2
=4
for x = 0
 π π
then, the number of points of discontinuity of f in  − ,  is
3 3
(A) 0
(B) 3
(C) 2
(D) 4
Q.44
Q.47
If y =
(A)
(a − x) a − x − (b − x) x − b
a −x + x−b
x + (a + b)
(a − x) (x − b)
(B)
then
dy
wherever it is defined is equal to :
dx
2 x − (a + b)
2 (a − x) (x − b)
(C) −
(a + b )
2 (a − x) (x − b)
(D)
2 x + (a + b)
2 (a − x) (x − b)
2
Q.48
If y is a function of x then
d y
dy
= 0 . If x is a function of y then the equation becomes :
2 +y
dx
dx
d2 x
dx
d2 x
+x
= 0 (B)
+y
(A)
dy
d y2
d y2
3
 dx
  =0
 dy
d2 x
(C)
−y
d y2
2
 dx
  =0
 dy
d2 x
(D)
−x
d y2
2
 dx
  =0
 dy
A function f (x) satisfies the condition, f (x) = f ′ (x) + f ′′ (x) + f ′′′ (x) + ...... ∞ where f (x) is a differentiable
function indefinitely and dash denotes the order of derivative . If f (0) = 1, then f (x) is :
(A) ex/2
(B) ex
(C) e2x
(D) e4x
cos 6x + 6 cos 4 x + 15 cos 2 x + 10
dy
Q.50 If y =
, then
=
cos 5x + 5 cos 3x + 10 cos x
dx
(A) 2 sinx + cosx
(B) –2sinx
(C) cos2x
(D) sin2x
Q.49
13 of 18
then :
3
d 2 x  dy 
d 2y


Q.51 If
+ 2 = K then the value of K is equal to
dy 2  dx 
dx
(A) 1
(B) –1
(C) 2
(D) 0
98930 58881
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Q.38
(C) f ′ (x) is not defined at x = sin 8
(D) f ′ (sin 8) < 0
A function f, defined for all positive real numbers, satisfies the equation f(x2) = x3 for every x > 0 . Then the value
of f ′ (4) =
(A) 12
(B) 3
(C) 3/2
(D) cannot be determined
TEKO CLASSES, MATHS BY SUHAAG SIR PH: (0755)- 32 00 000,
Q.37
(
(A)
2
x (1 − x)
1
2
2
x (1− x)
Let
 e
y = f(x) = 

 0
(D) π
1
if x ≠ 0
x2
if x = 0
(A)
(B)
(C)
1
(D)
1
1
 A + m  n − A  m+ n  A − m  n + A  m − n
m− n 
.  x n−A 
. x A − m
Q.54 Diffrential coefficient of  x











(C) – 1
(D) xAmn
x + h f ( x) − 2 h f ( h )
is equal to
Let f (x) be diffrentiable at x = h then Lim
x→ h
x−h
(A) f(h) + 2hf '(h)
(B) 2 f(h) + hf '(h)
(C) hf(h) + 2f '(h)
(D) hf(h) – 2f '(h)
(A) 1
Q.55
Q.56
Q.57
(B) 0
d 3y
Limit+
x→0
Q.60
Q.61
Q.62
Q.63
Q.64
(D) 24 a2 (ax + b)

x
x
 a arc tan
 has the value equal to
− b arc tan
a
b 
x x 
1
(a 2 − b 2 )
(C)
6a 2 b 2
(B) 0
a 2 − b2
(D)
3a 2 b2
x
Let f (x) be defined for all x > 0 & be continuous. Let f(x) satisfy f   = f ( x ) − f ( y) for all x, y & f(e) = 1.
y
Then :
 
(C) x.f(x)→1 as x→ 0 (D) f(x) = ln x
(B) f   → 0 as x → 0
 x
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
4
2
x − x +1
dy
If y = 2
and
= ax + b then the value of a + b is equal to
dx
x + 3x + 1
5π
5π
5π
5π
(A) cot
(B) cot
(C) tan
(D) tan
8
12
12
8
Suppose that h (x) = f (x)·g(x) and F(x) = f (g ( x) ) , where f (2) = 3 ; g(2) = 5 ; g'(2) = 4 ;
f '(2) = –2 and f '(5) = 11, then
(A) F'(2) = 11 h'(2)
(B) F'(2) = 22h'(2)
(C) F'(2) = 44 h'(2)
(D) none
Let f (x) = x3 + 8x + 3
which one of the properties of the derivative enables you to conclude that f (x) has an inverse?
(A) f ' (x) is a polynomial of even degree.
(B) f ' (x) is self inverse.
(C) domain of f ' (x) is the range of f ' (x).
(D) f ' (x) is always positive.
Which one of the following statements is NOT CORRECT ?
(A) The derivative of a diffrentiable periodic function is a periodic function with the same period.
(B) If f (x) and g (x) both are defined on the entire number line and are aperiodic then the function F(x) = f (x) .
g (x) can not be periodic.
(C) Derivative of an even differentiable function is an odd function and derivative of an odd differentiable function
is an even function.
(D) Every function f (x) can be represented as the sum of an even and an odd function
Select the correct alternatives : (More than one are correct)
(A) f(x) is bounded
Q.59
b g
If y = at2 + 2bt + c and t = ax2 + 2bx + c, then
equals
dx 3
2
2
(A) 24 a (at + b)
(B) 24 a (ax + b)
(C) 24 a (at + b)2
a−b
(A)
3
Q.58
w.r.t. x is
1
If y = tan x tan 2x tan 3x then
dy
has the value equal to :
dx
98930 58881
Then which of the following can best represent the graph of y = f(x) ?
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
(C) −
(B) zero
−
Q.53
)
−1
−1
If f(x) = 2 sin 1 − x + sin 2 x (1 − x) where x ∈  0 ,  then f ' (x) has the value equal to
14 of 18
Q.52
Q.67
Q.68
Q.69
2
2
dy
= (A) 2 ln x . x x (B) (2 ln x + 1). x x
dx
dy
Let y = x + x + x + ...... ∞ then
=
dx
1
1
x
(B)
(C)
(A)
2y − 1
x + 2y
1 + 4x
dy
If 2x + 2y = 2x + y then
has the value equal to :
dx
1
2y
(A) − x
(B)
(C) 1 − 2y
x
1
−
2
2
2
If y = x x then
The functions
u = ex sin x ;
v = ex cos x
dv
du
(A) v
−u
= u2 + v2
dx
dx
x −2 x −1
(C) (2 ln x + 1). x x
(D)
(D)
(C)
(D) x x
15 of 18
2 +1
(
(2
)
− 1)
2x 1 − 2y
2
y
x
d 2v
= − 2u
dx 2
(D) none of these
Q.70
Let f (x) =
Q.71
(A) f ′ (10) = 1
(B) f ′ (3/2) = − 1
(C) domain of f (x) is x ≥ 1 (D) none
Two functions f & g have first & second derivatives at x = 0 & satisfy the relations,
x −1 −1
. ln ex2
y
2x + y
satisfy the equation :
d2u
(B)
= 2v
dx2
2 + 1
98930 58881
Q.66
. x then :
2
, f ′ (0) = 2 g ′ (0) = 4g (0) , g ′′ (0) = 5 f ′′ (0) = 6 f(0) = 3 then :
g(0)
15
f (x)
(A) if h(x) =
then h ′ (0) =
(B) if k(x) = f(x) . g(x) sin x then k ′ (0) = 2
4
g(x)
1
g′ (x)
(C) Limit
=
(D) none
x→0
2
f ′ (x)
f(0) =
Q.72
If y = x (Anx )
(
An ( An x )
dy
is equal to :
dx
+ 2 An x An (An x )
, then
y
A n x An x − 1
x
y
((ln x)2 + 2 ln (ln x))
(C)
x An x
(A)
Q.1
A
Q.6
B
Q.11 B
Q.16 D
Q.21 A
Q.26 C
Q.31 D
Q.36 D
Q.41 C
Q.46 C
Q.51 D
Q.56 D
Q.61 B
Q.64 A,B,C
Q.68 A,B,C,D
Q.2
Q.7
Q.12
Q.17
Q.22
Q.27
Q.32
Q.37
Q.42
Q.47
Q.52
Q.57
Q.62
Q.65
Q.69
C
B
D
A
B
C
D
B
C
B
B
D
D
A,C
A,B,C
)
y
(ln x)ln (ln x) (2 ln (ln x) + 1)
x
y An y
(D)
(2 ln (ln x) + 1)
x An x
(B)
ANSWER KEY
Q.3
Q.8
Q.13
Q.18
Q.23
Q.28
Q.33
Q.38
Q.43
Q.48
Q.53
Q.58
Q.63
Q.66
Q.70
B
C
B
B
B
D
D
D
D
C
C
D
B
C,D
A,B
Q.4
Q.9
Q.14
Q.19
Q.24
Q.29
Q.34
Q.39
Q.44
Q.49
Q.54
Q.59
D
A
D
C
B
D
B
A
D
A
B
A
Q.5
Q.10
Q.15
Q.20
Q.25
Q.30
Q.35
Q.40
Q.45
Q.50
Q.55
Q.60
B
C
C
C
C
C
D
C
B
B
A
B
Q.67
Q.71
A,C,D
A,B,C
Q.72
B,D
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Q.65
(A) 3 sec2 3x tan x tan 2x + sec2 x tan 2x tan 3x + 2 sec2 2x tan 3x tan x
(B) 2y (cosec 2x + 2 cosec 4x + 3 cosec 6x)
(C) 3 sec2 3x − 2 sec2 2x − sec2 x
(D) sec2 x + 2 sec2 2x + 3 sec2 3x
dy
x
− x
If y = e + e
then
equals
dx
1
1
e x − e− x
e x − e− x
y2 − 4
y2 + 4
(B)
(A)
(C)
(D)
2 x
2 x
2 x
2x
16 of 18
EXERCISE -1
Part : (A) Only one correct option
dy
2
2 x 2 − 1 and y = f(x ) then dx at x = 1 is
(B) 1
(C) – 2
(A) 2
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2.
If y = x
x2
then
(A) 2 An x.
3.
4.
If f(x) =
1
(A)
2
2
2
(C) (2 An x + 1).
xx
2 + 1
2
(D) x x + 1 . An ex2
, then f′(0).
x
1
2
then
x
b+
xx
(B) (2 An x + 1).
(B) –
If y =
(A)
dy
=
dx
x

tan −1  sin 
2

e
a+
5.
xx
(D) –1
(C) 1
(D) –1
dy
=
dx
x
a +.......... .....
a
ab + 2 ay
(B)
b
ab + 2 by
(C)
a
ab + 2 by
(D)
b
ab + 2 ay
d2 F
Let f(x) = sin x; g(x) = x & h(x) = loge x & F(x) = h[g(f(x))] then
is equal to:
dx 2
2
(A) 2 cosec3 x (B) 2 cot (x2) − 4x2 cosec 2 (x2)
6.
7.
(C) 2x cot x2
(D) − 2 cosec2 x
dy
at x = 0 is
If y = (1 + x) (1 + x2) (1 + x 4) .....(1 + x2n), then
dx
(A) –1
(B) 1
(C) 0
(D) 2n
dy
1
If y = sin−1  x 1 − x + x 1 − x 2  and
=
+ p, then p =


dx
2 x (1 − x)
(A) 0
8.
9.
10.
12.
13.
14.
(C) sin−1 x
1− x

y

If x 2 + y 2 = et where t = sin–1 
2
 x +y2

(D)

dy

 then dx :

x−y
x+y
(B)
(A)
x
4
x −1
(B)
x2
x4 − 1
The differential coefficient of sin−1
(C) 0
t
1+ t
2
w.r.t. cos−1
(B) t
(C)
1
1− x 2
x−y
(D) 2x + y
2x + y
x+y
(C)
x−y
x−y
2
2
dy
x −1
x +1
If y = sin−1 2
+ sec−1 2
, x > 1 then
is equal to:
dx
x +1
x −1
(A)
(A) 1
11.
1
(B)
(D) 1
1
1 + t2
1
is:
(D) none
1 + t2
 tan − 1 x 
−1

− 1  w.r.t. tan x is:
1
tan
x
+


Differentiation of 
−1
1
1



(A) 
(B) − 1
(C)
(D)
−1
2
(1 + tan −1 x )2
 1 + tan x 
1 + tan −1 x
Let f(x) be a polynomial in x. Then the second derivative of f(ex), is:
(A) f ′′ (ex). ex + f ′ (ex)
(B) f ′′ (ex). e2x + f ′ (ex). e2x
(C) f ′′ (ex) e2x
(D) f ′′ (ex). e2x + f ′ (ex). ex
f g h
f ′ g′ h′
If f(x), g(x), h(x) are polynomials in x of degree 2 and F(x) =
, then F′(x) is equal to
f ′′ g′′ h′′
(A) 1
(B) 0
(C) –1
(D) f(x) . g(x) . h(x)
y y1 y 2
(
If y = sin mx then the value of y 3
y6
y4
y7
)
y 5 (where settings of y shows the order of derivative) is:
y8
98930 58881
If f′(x) =
TEKO CLASSES, MATHS BY SUHAAG SIR PH: (0755)- 32 00 000,
1.
(B) dependent of x but independent of m
(D) independent of m & x.
f (5 + t ) − f (5 − t )
=
2t
15.
If f ′ (5) = 7 then Limit
t →0
16.
(A) 0
(B) 3.5
(C) 7
(D) 14
Let ef(x) = ln x. If g(x) is the inverse function of f(x) then g′ (x) equals to:
(A) ex
(C) ex+e
(B) ex + x
x
17 of 18
(A) independent of x but dependent on m
(C) dependent on both m & x
(D) ex + ln x
If u = ax + b then
(A)
18.
(C) an
dy
 2x − 1
=
 & f ′ (x) = sin x then
2
dx
 x + 1
dn
[f(u)]
du n
(D) a−n
(C)
1 + x − x2
(1 + x )
2
2
1 − x + x2
(1 + x )
2
(B)
 2x − 1

 x 2 + 1
(D) none
sin 
2
(
) sin  2x − 1
 x + 1
(1 + x )
2 1 + x − x2
 2x − 1

 x 2 + 1
sin 
2
2
2
 d   3 d 2y 
  y . 2  equals:
 dx  
dx 
If y2 = P(x), is a polynomial of degree 3, then 2 
(A) P ′′′ (x) + P ′ (x)
(B) P ′′ (x). P ′′′ (x)
(C) P (x). P ′′′ (x)
Part : (B) May have more than one options correct
20.
dn
[f(u)]
dx n
If y = f 
(A)
19.
dn
[f(u)]
du n
d
[f(ax + b)] is equal to:
dx n
dn
(B) a n [f(u)]
du
(D) a constant
Two functions f & g have first & second derivatives at x = 0 & satisfy the relations,
2
, f ′ (0) = 2 g ′ (0) = 4g (0), g ′′ (0) = 5 f ′′ (0) = 6 f(0) = 3 then:
g(0)
15
f (x)
(A) if h(x) =
then h ′ (0) =
(B) if k(x) = f(x). g(x) sin x then k ′ (0) = 2
4
g(x)
g′ (x)
1
(C) Limit
=
(D) none
x →0
f ′ (x)
2
f(0) =
21.
If fn (x) = e
fn − 1 ( x )
for all n ∈ N and fo (x) = x, then
d
{f (x)} is equal to:
dx n
22.
d
{f
(x)}
(B) fn (x). fn − 1 (x)
dx n − 1
(D) none of these
(C) fn (x). f n − 1 (x)........ f2 (x). f1 (x)
If f is twice differentiable such that f′′(x) = –f(x) and f′(x) = g(x). If h(x) is a twice differentiable function such that
h′(x) = [f(x)]2 + [g(x)]2 . If h(0) = 2, h(1) = 4, then the equation y = h(x) represents:
(A)
a curve of degree 2
(B)
a curve passing through the origin
(C)
a straight line with slope 2
(D)
a straight line with y intercept equal to 2.
23.
Given f(x) = −
(A) fn (x).
24.
25.
x3
+ x2 sin 1.5 a − x sin a. sin 2a − 5 sin–1 (a2 − 8a + 17) then:
3
(A) f′(x) = – x 2 + 2x sin6 – sin4 sin8
(B) f ′ (sin 8) > 0
(C) f ′ (x) is not defined at x = sin 8
(D) f ′ (sin 8) < 0
If f(x) = x3 + x2 f′(1) + xf′′(3) for all x ∈ R then
(A) f(0) + f(2) = f(1)
(B) f(0) + f(3) = 0
(C) f(1) + f(3) = f(2)
(D) none of these
If f(x) = (ax + b) sin x + (cx + d) cos x, then the values of a, b, c and d such that f′(x) = x cos x for all x are
(A) a = d = 1
(B) b = 0
(C) c = 0
(D) b = c
EXERCISE -2
d2 y
dy
+ n2 y = 0, where n2 = p2 + k2.
dt
dt
Evaluate the following limits using L′′ hospitale rule as otherwise
2.
Limit log tan2 x (tan2 2 x)
1.
If y = A e − kt cos (p t + c) then prove that
2
+2k
x →0
3.
If f (x) =
( x−a) 4
( x−a)3 1
( x−b) 4
( x−b )3 1
( x−c )
4
( x−c )
3
1
then f ′ (x) = λ.
( x−a) 4
( x−a )2 1
( x−b) 4
( x−b)2 1
4
( x−c )2 1
( x−c )
. Find the value of λ.
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17.
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n
where t is a parameter, then prove that
5.
If sin y = x sin (a + y), show that
d3 y
dx
3
=
8.b
18 of 18
If x = a
27a 3 .t 7
sina
dy
=
2 .
dx 1 − 2xcosa+x
F" f " g" 2c
F′′′ f ′′′ g′′′
If F(x) = f(x). g(x) & f ′(x). g ′(x) = c, prove that F = f + g + f g & F = f + g .
If α be a repeated root of a quadratic equation f(x) = 0 & A(x), B(x), C(x) be the polynomials of degree 3, 4 & 5
A( x ) B( x ) C( x )
respectively, then show that
3/2
can be reduced to the form R2/3 =
1
Show that R =
9.
d y


 dx 2 


Also show that, if x = a sin 2θ (1 + cos 2θ) & y = a cos 2θ (1 − cos
4 a cos 3 θ.
Differentiate the following functions with respect to x.
2
d y
2
dx 2
(i)
x2. An x. ex
(i)
sinx − xcos x
x sinx + cos x
2
3
1
+
8.
2
3
.
d x


 dy 2 


2θ) then the value of R equals to
2
(iii)

−1 1 − cos x
tan  tan
1 + cos x

9.
20.
C
ABC
(iii)
1
x
sec2
2
2




Exercise # 1
1. A 2. C 3. A 4. D 5. D
12. D 13. B 14. D 15. C 16. C
21. AC 22. CD 23. AD 24. ABC
6. B 7. D
17. C 18. B
25. ABC
8.
19.
B
C
10.
A
11.
Exercise # 2
2. 1
3. 3
9. (i) ex x (2 An x + 1 + x An x)
(ii)
x2
( xsinx + cos x)2
for 32 Yrs. Que. of IIT-JEE
&
8 Yrs. Que. of AIEEE
we have distributed already a book
C
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2

1 +  dy  
  dx  
A(α ) B(α ) C(α )
is divisible by f(x), where dash denotes the derivative.
A ' (α ) B' (α) C' (α )
TEKO CLASSES, MATHS BY SUHAAG SIR PH: (0755)- 32 00 000,
7.
&y=b
t2,
4.
6.
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t3