If $\tan ^{-1} y-y+x=0$ then $\frac{d^2 y}{d x^2}$ is equal to

$\frac{2\left(1+y^2\right)}{y^4}$

If $x = a\sin {\rm{\theta }}\,$ and $y = b\cos {\rm{\theta }},$ then $\frac{{{d^2}y}}{{d{x^2}}}$ is equal to

$ - \frac{b}{{{a^2}}}{\sec ^3}{\rm{\theta }}$

$\frac{a}{{{b^2}}}{\sec ^2}{\rm{\theta }}$

$ - \frac{b}{a}{\sec ^2}{\rm{\theta }}$

$\frac{b}{{{a^2}}}{\sec ^3}{\rm{\theta }}$

If $y = {e^{1 + {{\log }_{\rm{e}}}x}}$, then the value of $\frac{{dy}}{{dx}}$ is equal to

${\log _e}x{e^{{{\log }_e}^x}}$

The function $f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{{\tan x}}{x},x \ne 0}\\ {1,x = 0} \end{array}} \right.$, is

Continuous but not differentiable at $x = 0$

If ${2^x} + {2^y} = {2^{x + y}}$, then $\frac{{dy}}{{dx}}$ is equal to

$\frac{{{2^x} + {2^y}}}{{{2^x} - {2^y}}}$

$\frac{{{2^x} + {2^y}}}{{1 + {2^{x + y}}}}$

$\frac{{{2^{x + y}} - {2^x}}}{{{2^y}}}$

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JEE • MATHEMATICS • DIFFERENTIATION

Differentiation practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 31-60 of 932 questions

31 Medium

31 of 932

Analyzing +4 | -1

The derivative of

{\tan ^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right)

with respect to

{\cos ^{ - 1}}\sqrt {1 - {x^2}}

View answer and explanation

Correct option: D

\frac{{2\sqrt {1 - {x^2}} }}{{1 + {x^2}}}

Option D (

\frac{{2\sqrt {1 - {x^2}} }}{{1 + {x^2}}}

) is correct because it results from correctly differentiating both functions and then dividing the derivatives.

32 Easy

32 of 932

Analyzing +4 | -1

f\left( x \right) = \frac{{x - 1}}{4} + \frac{{{{\left( {x - 1} \right)}^3}}}{{12}} + \frac{{{{\left( {x - 1} \right)}^5}}}{{20}} + \frac{{{{\left( {x - 1} \right)}^7}}}{{28}} + \ldots ,

where

0 < x < 2

, then

f'\left( x \right)

is equal to

View answer and explanation

Correct option: A

\frac{1}{{4x\left( {2 - x} \right)}}

Option A (

\frac{1}{{4x\left( {2 - x} \right)}}

) is the correct derivative, obtained by recognizing the series as a logarithmic form and then differentiating.

33 Easy

33 of 932

Application +4 | -1

f

is one-one and satisfies

f^{\prime}(x)=\sqrt{1-(f(x))^2}

then

\left(f^{-1}\right)^{\prime}(x)

View answer and explanation

Correct option: A

is equal to

\frac{1}{\sqrt{1-x^2}}

Option A (is equal to

\frac{1}{\sqrt{1-x^2}}

) is correct because it is the derivative of the inverse function using the given condition.

34 Easy

34 of 932

Application +4 | -1

The derivative of

f(x)=\left|x^{2}-x\right|

x=2

View answer and explanation

Correct option: C

Option C (3) is correct because the derivative of the function at x=2 is calculated as 3 after considering the absolute value.

35 Easy

35 of 932

Application +4 | -1

Let

f\left( x \right) = {\left( {x - 7} \right)^2}{\left( {x - 2} \right)^7},x \in \left[ {2,7} \right].

The value of

\theta \in \left( {2,7} \right)

such that

f'\left( \theta \right) = 0

is equal to

View answer and explanation

Correct option: B

\frac{{53}}{9}

Option B (

\frac{{53}}{9}

) is correct because it is the value of theta that makes the derivative of the function equal to zero within the given interval.

36 Easy

36 of 932

Analyzing +4 | -1

f(x)=\cos x \cos 2 x \cos 4 x \cos 8 x \cos 16 x

, then

f^{\prime}\left(\frac{\pi}{4}\right)

View answer and explanation

Correct option: A

\sqrt{2}

Option D (All the above) is correct because

f'(2) = f'(3) = 0

and

f'(\frac{1}{2}) = \frac{16}{5}

, satisfying all the given conditions.

37 Easy

37 of 932

Application +4 | -1

Let

f(\cdot)

and

g(\cdot)

be differentiable functions on

(-\infty

\infty)

and let

f^{\prime}(\cdot)

and

g^{\prime}(\cdot)

denote derivatives of

f(\cdot)

and

g(\cdot)

respectively. If

f(0)=\frac{1}{2}, g(0)=\frac{1}{3}, f^{\prime}(0)=1

and

g^{\prime}(0)=2

, then the value of

\lim \limits_{{x \rightarrow 0}} \frac{2 f\left(2 x^2+3 x\right)-1}{3 g(x)-1}

View answer and explanation

Correct option: A

Option A (1) is correct because the limit is evaluated using L'Hopital's rule and the given derivative values.

38 Easy

38 of 932

Application +4 | -1

y = {\sin ^{ - 1}}\sqrt {1 - x}

, then

\frac{{dy}}{{dx}}

is equal to

View answer and explanation

Correct option: D

\frac{{ - 1}}{{2\sqrt x \sqrt {1 - x} }}

Option D (

\frac{{ - 1}}{{2\sqrt x \sqrt {1 - x} }}

) is correct because it is the result of applying the chain rule correctly when differentiating the inverse sine function.

39 Easy

39 of 932

Application +4 | -1

Let

f: R \rightarrow R

be a function given by

f(x + y) = f(x)f(y)\quad \mathrm{ for}\,\,\mathrm{ all}\,\,\mathrm{ }x,y \in R

.
If

f(x) = 1 + xg(x) + {x^2}g(x)\phi (x)\,\,\,\,\mathrm{such}\,\,\,\mathrm{that}\,\,\,\,\lim\limits_{x \to 0} g(x) = a\,\,\,\mathrm{and}\,\,\lim\limits_{x \to 0} \phi (x) = b\mathrm{,}\,\,\,\mathrm{then}\,\,\,f'(x)\,\,\,\mathrm{is}\,\,\,\mathrm{equal}\,\,\,\mathrm{to}

View answer and explanation

Correct option: B

a f (x)

Option B is correct because using the given condition

f(x) = 1 + xg(x) + {x^2}g(x)\phi (x)

and the functional equation, we find

f'(x) = a f (x)

40 Medium

40 of 932

Analyzing +4 | -1

f

is defined in

[-5,5]

\begin{aligned} f(x) & =x \text { if } x \text { is rational } \\ & =-x \text { if } x \text { is irrational. Then } \end{aligned}

View answer and explanation

Correct option: B

f(x)

is discontinuous at every

x

, except

x=0

Option B is correct because the function is discontinuous at every x except x=0, where it is continuous.

41 Medium

41 of 932

Application +4 | -1

y=f\left(\frac{2 x-1}{x^2+1}\right)

and

f^{\prime}(x)=\sin x^2

, then

\frac{d y}{d x}=

View answer and explanation

Correct option: D

\frac{-2 x^2+2 x+2}{\left(x^2+1\right)^2} \sin \left(\frac{2 x-1}{x^2+1}\right)^2

Option D (

\frac{{-2 x^2+2 x+2}}{{(x^2+1)^2}}

\sin (

\frac{{2 x-1}}{{x^2+1}}

)²) is correct because it is the result of applying the chain rule to find the derivative of the composite function.

42 Easy

42 of 932

Application +4 | -1

Let

g(x)

be the inverse of

f(x)

and

f^{\prime}(x)=\frac{1}{1+x^5}

then

\frac{d^2}{d x^2}(g(x))

is equal to

View answer and explanation

Correct option: B

5(g(x))^4\left(1+(g(x))^5\right.

Option B (

5(g(x))^4(1+(g(x))^5)

) is correct because it is the second derivative of the inverse function

g(x)

43 Medium

43 of 932

Analyzing +4 | -1

Let

g(x)=\log f(x)

, where

f(x)

is a twice differentiable positive function on

(0

\infty)

such that

f(x+1)=x f(x)

. Then, for

N=1,2,3, \ldots, g^{\prime \prime}\left(N+\frac{1}{2}\right)-g^{\prime \prime}\left(\frac{1}{2}\right)

is equal to

View answer and explanation

Correct option: A

-4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N-1)^2}\right\}

Option A (

-4\left\{1+\frac{{1}}{{9}}+\frac{{1}}{{25}}+\ldots+\frac{{1}}{{{(2 N-1)}^2}}\right\}

) is correct because it is the result obtained after substituting and simplifying the given functional equation and differentiating twice.

44 Medium

44 of 932

Application +4 | -1

The derivative of

{\sin ^{ - 1}}\left( {\frac{{\sqrt {1 + x} + \sqrt {1 - x} }}{2}} \right)

with respect to

x

View answer and explanation

Correct option: A

- \frac{1}{{2\sqrt {1 - {x^2}} }}

Option A,

- \frac{1}{{2\sqrt {1 - {x^2}} }}

, is correct because it is the result of differentiating the given inverse trigonometric function after a suitable substitution to simplify the expression.

45 Easy

45 of 932

Application +4 | -1

\tan ^{-1} y-y+x=0

then

\frac{d^2 y}{d x^2}

is equal to

View answer and explanation

Correct option: A

\frac{-2\left(1+y^2\right)}{y^5}

Option A (

\frac{-2(1+y^2)}{y^5}

) is correct because it represents the second derivative of y with respect to x, obtained through implicit differentiation.

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46 Medium

46 of 932

Application +4 | -1

\frac{d}{d x}\left(\cos ^{-1} \frac{x-x^{-1}}{x+x^{-1}}\right)=

View answer and explanation

Correct option: B

\frac{-2}{1+x^2}

Option B is correct because the expression simplifies to an inverse tangent function, and its derivative is -2/(1+x²).

47 Medium

47 of 932

Application +4 | -1

Let

y=\log _e\left(\frac{1-x^2}{1+x^2}\right),-1 < x < 1

. Then at

x=\frac{1}{2}

, the value of

225\left(y^{\prime}-y^{\prime \prime}\right)

is equal to

View answer and explanation

Correct option: D

736

The correct answer is D (736) because it is the result of correctly calculating the first and second derivatives and substituting x=1/2 into the expression.

48 Medium

48 of 932

Application +4 | -1

x = \frac{{1 - \sqrt y }}{{1 + \sqrt y }} \Rightarrow \frac{{dy}}{{dx}}

is equal to

View answer and explanation

Correct option: B

\frac{{4\left( {x - 1} \right)}}{{{{\left( {1 + x} \right)}^3}}}

Option B (

\frac{{4\left( {x - 1} \right)}}{{{{\left( {1 + x} \right)}^3}}}

) is correct because it is the result of correctly applying componendo and dividendo, followed by differentiation.

49 Medium

49 of 932

Application +4 | -1

y = {\log _a}x + {\log _x}a + {\log _x}x + {\log _a}a

, then

\frac{{dy}}{{dx}}

is equal to

View answer and explanation

Correct option: D

\frac{1}{{x\log a}} - \frac{{\log a}}{{x{{\left( {\log x} \right)}^2}}}

Option D is correct because the derivative is found using logarithm properties and the chain rule.

50 Medium

50 of 932

Application +4 | -1

Suppose for a differentiable function

h, h(0)=0

h(1)=1

and

h^{\prime}(0)=h^{\prime}(1)=2

. If

g(x)=h\left(e^x\right) e^{h(x)}

, then

g^{\prime}(0)

is equal to:

View answer and explanation

Correct option: D

Option D (4) is correct because it is the value of the derivative of g(x) evaluated at x=0, calculated using the chain rule and product rule.

51 Medium

51 of 932

Application +4 | -1

x = a\sin \mathrm{\theta }\,

and

y = b\cos \mathrm{\theta },

then

\frac{{{d^2}y}}{{d{x^2}}}

is equal to

View answer and explanation

Correct option: D

- \frac{b}{{{a^2}}}{\sec ^3}\mathrm{\theta }

Option D (

- \frac{b}{{{{a^2}}}}\sec ^3\mathrm{\theta }

) is correct because it is the result of differentiating the given parametric equations twice with respect to x.

52 Medium

52 of 932

Application +4 | -1

y = {e^{1 + {{\log }_\mathrm{e}}x}}

, then the value of

\frac{{dy}}{{dx}}

is equal to

View answer and explanation

Correct option: A

e

The correct answer is

e

. The function simplifies to

y = ex

, and its derivative is

e

53 Easy

53 of 932

Analyzing +4 | -1

f\left( x \right) = {\left( {1 - x} \right)^n}

, then the value of

f\left( 0 \right) + f'\left( 0 \right) + \frac{{f''\left( 0 \right)}}{{2!}} + \ldots \frac{{{f^n}\left( 0 \right)}}{{n!}}

, is

View answer and explanation

Correct option: B

Option B (0) is correct because the given expression is the binomial expansion of

(1-1)^n

, which equals 0.

54 Easy

54 of 932

Application +4 | -1

The tangent line at

(2,4)

to the curve

y=

x^3-3 x+2

meets the

x

-axis at

View answer and explanation

Correct option: D

\left(\frac{14}{9}, 0\right)

Option D

\left(\frac{{14}}{{9}}, 0\right)

is correct because it represents the x-intercept of the tangent line to the given curve at the point (2,4).

55 Medium

55 of 932

Analyzing +4 | -1

The function

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{\tan x}}{x},x \ne 0}\\ {1,x = 0} \end{array}} \right.

, is

View answer and explanation

Correct option: C

Continuous and differentiable at

x = 0

Option C (Continuous and differentiable at

x = 0

) is correct because the function satisfies both continuity and differentiability conditions at the specified point.

56 Medium

56 of 932

Analyzing +4 | -1

{2^x} + {2^y} = {2^{x + y}}

, then

\frac{{dy}}{{dx}}

is equal to

View answer and explanation

Correct option: C

{2^{x - y}}\left( {\frac{{{2^y} - 1}}{{1 - {2^x}}}} \right)

Option C (

{2^{x - y}}\left( {\frac{{{{2^y}} - 1}}{{1 - {{2^x}}}}} \right)

) is correct because it is the result of implicit differentiation and algebraic manipulation of the given equation.

57 Easy

57 of 932

Application +4 | -1

x = \sin t,y = \cos pt,

, then

View answer and explanation

Correct option: D

\left( {1 - {x^2}} \right){y_2} - x{y_1} + {p^2}y = 0

Option D (

\left( {1 - {x^2}} \right){y_2} - x{y_1} + {p^2}y = 0

\) is correct because it is the differential equation obtained after differentiating the given parametric equations and eliminating the parameter

t

58 Easy

58 of 932

Application +4 | -1

g

is the inverse of function

f

and

f^{\prime}(x)=\sin x

, then

g^{\prime}(x)=

View answer and explanation

Correct option: C

\frac{1}{\sin \,\{g(x)\}}

Option C (

\frac{1}{\sin \\{g(x)\\}}

) is correct because it is derived from differentiating the composite function

f(g(x)) = x

using the chain rule.

59 Medium

59 of 932

Application +4 | -1

f(x)=(1-x) n, n \in N

then

f(0)+\frac{f^{\prime}(0)}{1 !}+\frac{f^{\prime \prime}(0)}{2 !}+\ldots . .+\frac{f^n(0)}{n !}=

View answer and explanation

Correct option: B

Option C (0) is correct because the given expression is the binomial expansion of (1-1)^n, which equals 0.

60 Easy

60 of 932

Application +4 | -1

The differential coefficient of

f\left(\log _{e} x\right)

w.r.t.

x

, where

f(x)=\log _{e} x

View answer and explanation

Correct option: C

\left(x \log _{e} x\right)^{-1}

Option C (

(x \log_e x)^{-1}

) is correct because it is the result of applying the chain rule to find the derivative of the composite function.

Revision notes and key takeaways

Review DIFFERENTIATION in your textbook or class notes, then use this set to check your understanding.

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