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 · Medium

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Practice format and chapter quiz

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 1-30 of 402 questions

1 Medium

1 of 402

Analyzing +4 | -1

Let

f(x)=[x] \sin (\pi x)

where

[x]

denotes the integer part of

x

. Then at

x=k, k

being an integer, the left derivative of

f(x)

View answer and explanation

Correct option: A

(-1)^k(k-1) \pi

Option A (

(-1)^k(k-1) \pi

) is correct because it is the result of calculating the left derivative of the function at x=k using the limit definition.

2 Medium

2 of 402

Application +4 | -1

Let

f(x)=\log \sin x^2, 0 < x < \pi / 2

then

f^{\prime \prime}(\sqrt{\pi} / 2)

is equal to

View answer and explanation

Correct option: C

2(1-\pi)

Option C (

2(1-\pi)

) is correct because it is the value of the second derivative of the given function evaluated at

x = \sqrt{\pi}/2

3 Medium

3 of 402

Analyzing +4 | -1

Let

f

be any function defined on

R

and let it satisfy the condition

\left| {f(x) - f(y)\left| {{\mkern 1mu} \le {\mkern 1mu} } \right|{{\left( {x - y} \right)}^2}} \right|,{\mkern 1mu} {\mkern 1mu} \forall \;x,y \in R

f(0) = 1

, then

View answer and explanation

Correct option: D

f(x) > 0,\,\,\forall x \in R

Option D is correct. Given the condition on

f(x)

, it can be shown that

f(x)

is a constant function, and since

f(0) = 1

f(x) = 1

for all

x

, thus

f(x) > 0

for all

x \in R

4 Medium

4 of 402

Analyzing +4 | -1

A function

f: R \rightarrow R

is differentiable and satisfies the equation

f\left(\frac{1}{n}\right)=0

for all integers

n \geq 1

, then

View answer and explanation

Correct option: B

f(0)=0=f^{\prime}(0)

Option B (

f(0)=0=f^{\prime}(0)

) is correct because the differentiability of

f

x=0

along with the given condition

f(1/n) = 0

forces both

f(0)

and

f'(0)

to be zero.

5 Medium

5 of 402

Application +4 | -1

f(x)=(2 x-3 \pi)^{25}+\frac{4}{3} x+\cos x

and

g(x)

is the inverse of

f(x)

, then find

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

x=2 \pi

View answer and explanation

Correct option: B

\frac{3}{7}

Option B (3/7) is correct because it is the derivative of the inverse function g(x) evaluated at x=2*pi, which is the reciprocal of the derivative of f(x) evaluated at f^-1(2*pi).

6 Medium

6 of 402

Application +4 | -1

y=\frac{e^{2 x} \cos x}{x \sin x}

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: A

\frac{e^{2 x}\left[(2 x-1) \cot x-x \operatorname{cosec}^{2} x\right]}{x^{2}}

Option A is correct because the derivative is found using logarithmic differentiation and simplifying the resulting expression.

7 Medium

7 of 402

Analyzing +4 | -1

y=y(x)

and it follows the relation

x \cos y+y \cos x=\pi

then

y^{\prime \prime}(0)=

View answer and explanation

Correct option: C

\pi

Option C (

\pi

) is correct because after differentiating the given equation twice and substituting x=0, we find that y''(0) =

\pi

8 Medium

8 of 402

Application +4 | -1

The derivative of

\left[ {\frac{{{e^x} + 1}}{{{e^x}}}} \right]

is equal to

View answer and explanation

Correct option: C

- \frac{1}{{{e^x}}}

Option C (

- \frac{1}{{{e^x}}}

) is correct because it is the derivative of the given function after simplification.

9 Medium

9 of 402

Application +4 | -1

f\left( x \right) = 10\cos x + \left( {13 + 2x} \right)\sin x

, then

f''\left( x \right) + f\left( x \right) =

View answer and explanation

Correct option: B

4\cos x

Option B (

4\cos x

) is correct because after finding the second derivative and adding it to the original function, the expression simplifies to

4\cos x

10 Medium

10 of 402

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.

11 Medium

11 of 402

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.

12 Medium

12 of 402

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.

13 Medium

13 of 402

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.

14 Medium

14 of 402

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.

15 Medium

15 of 402

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).

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16 Medium

16 of 402

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.

17 Medium

17 of 402

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.

18 Medium

18 of 402

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.

19 Medium

19 of 402

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.

20 Medium

20 of 402

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.

21 Medium

21 of 402

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

22 Medium

22 of 402

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.

23 Medium

23 of 402

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.

24 Medium

24 of 402

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.

25 Medium

25 of 402

Application +4 | -1

f^{\prime}(1)=2

and

g^{\prime}(\sqrt{2})=4

, then the derivative of

f(\tan x)

w.r.t.

g(\sec x)

x=\frac{\pi}{4}

View answer and explanation

Correct option: C

\sqrt{2} / 2

Option C (

\sqrt{2} / 2

) is correct because it is the result of differentiating

f(\tan x)

with respect to

g(\sec x)

and evaluating at

x = \frac{\pi}{4}

26 Medium

26 of 402

Analyzing +4 | -1

f(x)=x^{3}-x^{2} f^{\prime}(1)+x f^{\prime \prime}(2)-f^{\prime \prime}(3), x \in R

, then

View answer and explanation

Correct option: A

2 f(0)-f(1)+f(3)=f(2)

Option A (

2 f(0)-f(1)+f(3)=f(2)

) is correct because after finding the derivatives and solving for the constants, the equation holds true.

27 Medium

27 of 402

Application +4 | -1

f\left( {x + y} \right) = f\left( x \right)f\left( y \right)

for all

x,y \in R,f\left( 5 \right) = 2,f'\left( 0 \right) = 3

. Then

f'\left( 5 \right)

equals

View answer and explanation

Correct option: A

Option A (6) is correct because by using the given functional equation and derivative information, we can find f'(5).

28 Medium

28 of 402

Application +4 | -1

y = {\log _u}\left| {\cos 4x} \right| + \left| {\sin x} \right|

, where

u = \sec 2x

, then

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

x = - \frac{\pi }{6}

is equal to

View answer and explanation

Correct option: A

\frac{{ - 6\sqrt 3 }}{{{{\log }_e}2}} - \frac{{\sqrt 3 }}{2}

Option A (

\frac{{ - 6\sqrt 3 }}{{{{\log }_e}2}} - \frac{{\sqrt 3 }}{2}

) is correct because it is the result of differentiating the given function and substituting

x = -\frac{\pi}{6}

29 Medium

29 of 402

Application +4 | -1

y = f\left( {\frac{{2x + 3}}{{3 - 2x}}} \right)

and

f\left( x \right) = \sin (\log x),

then

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

is equal to

View answer and explanation

Correct option: A

\frac{{12}}{{9 - 4{x^2}}}\cos \left\{ {\log \left( {\frac{{2x + 3}}{{3 - 2x}}} \right)} \right\}

Option A (

\frac{{12}}{{9 - 4{x^2}}}\cos \left\{ {\log \left( {\frac{{2x + 3}}{{3 - 2x}}} \right)} \right\}

) is correct because it follows from applying the chain rule to differentiate the composite function.

30 Medium

30 of 402

Application +4 | -1

f(x)=\sqrt{x}(x \geq 0)

and

g(x)=1+x^2

, then

(f \circ g)^{\prime}(1)

is equal to

View answer and explanation

Correct option: C

\frac{1}{\sqrt{2}}

Option C (

\frac{1}{\sqrt{2}}

) is correct because, after finding

(f \circ g)(x) = \sqrt{1+x^2}

, we differentiate and evaluate at

x=1

, resulting in

\frac{1}{\sqrt{2}}

Revision notes and key takeaways

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

Key Takeaways

✓ This set contains 402 single-correct-answer MCQs on DIFFERENTIATION.
✓ Try each question before opening its answer. Read the explanation where one is available.
✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

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