If $x = \phi \left( t \right),y = {\rm{\psi }}\left( t \right),$ then $\frac{{{d^2}y}}{{d{x^2}}}$ is equal to

$\frac{{\phi '{\rm{\psi ''}} - {\rm{\psi '}}\phi ''}}{{{{\left( {\phi '} \right)}^2}}}$

$\frac{{\phi ''}}{{{\rm{\psi ''}}}}$

If $ y=\left(1+\frac{1}{x}\right)^{x} $, then $ \frac{d y}{d x} $ is equal to

$ \left(1+\frac{1}{x}\right)^{x}\left[\log \left(1+\frac{1}{x}\right)\right] $

$ \left(x+\frac{1}{x}\right)^{x}\left[\log (x-1)-\frac{x}{x+1}\right] $

If $f(x-y), f(x) f(y)$ and $f(x+y)$ are in A.P. for all $x, y \in R$ and $f(0)=0$, then

$f^{\prime}(-3)+f^{\prime}(3)=0$

If $y = {\sin ^{ - 1}}\sqrt {1 - x} $, then $\frac{{dy}}{{dx}}$ is equal to

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

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

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

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JEE • MATHEMATICS • DIFFERENTIATION

Differentiation practice · Easy

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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 500 questions

1 Easy

1 of 500

Application +4 | -1

f(x)=|x|^{|\tan x|}

, then

f^{\prime}\left(\frac{-\pi}{6}\right)

is equal to

View answer and explanation

Correct option: B

\left(\frac{\pi}{6}\right)^{1 / \sqrt{3}}\left\{\frac{-2 \sqrt{3}}{\pi}+\frac{4}{3} \log \frac{6}{\pi}\right\}

Option B is correct because it is the result of differentiating the given function and evaluating at the specified point, considering the absolute value.

2 Easy

2 of 500

Application +4 | -1

Let

x

and

y

be two real numbers such that

x>0

and

x y=1

. The minimum value of

x+y

View answer and explanation

Correct option: C

Option C (2) is correct because the minimum value of x+y, given xy=1 and x>0, is 2, which can be found using calculus or AM-GM inequality.

3 Easy

3 of 500

Analyzing +4 | -1

y = {\sec ^{ - 1}}\left[ {\mathrm{cosec}x} \right] + \mathrm{cose}{\mathrm{c}^{ - 1}}\left[ {\sec x} \right] + {\sin ^{ - 1}}[\cos x] + {\cos ^{ - 1}}[\sin x],

then

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

is equal to

View answer and explanation

Correct option: D

- 4

Option D (-4) is correct because after simplifying the expression using trigonometric identities, the derivative is found to be -4.

4 Easy

4 of 500

Application +4 | -1

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

is equal to

View answer and explanation

Correct option: A

\frac{3}{\sqrt{1-x^{2}}}

The correct answer is A (

\frac{3}{\sqrt{1-x^{2}}}

) because it is the derivative of the simplified form of the given function after using trigonometric substitution.

5 Easy

5 of 500

Application +4 | -1

z = y + f\left( v \right),

where

v = \left( {\frac{x}{y}} \right),

then

v\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}}

View answer and explanation

Correct option: B

Option B (1) is correct because after partial differentiation and substitution, the terms involving the function f' cancel out, leaving 1.

6 Easy

6 of 500

Application +4 | -1

The minimum value of

f(x)=|3-x|+

|2+x|+|5-x|

View answer and explanation

Correct option: B

Option B (7) is correct because it represents the minimum value of the given function, which occurs at x=3.

7 Easy

7 of 500

Application +4 | -1

x = \phi \left( t \right),y = \mathrm{\psi }\left( t \right),

then

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

is equal to

View answer and explanation

Correct option: B

\frac{{\phi '\psi '' - \psi '\phi ''}}{{{{\left( {\phi '} \right)}^3}}}

Option B (

\frac{{\phi '\psi '' - \psi '\phi ''}}{{{{\left( {\phi '} \right)}^3}}})

is correct because it represents the second derivative of y with respect to x when both x and y are parameterized by t.

8 Easy

8 of 500

Analyzing +4 | -1

f\left( x \right) = \frac{1}{{1 - x}},

then the derivative of the composite function

f\left[ {f\left\{ {f\left( x \right)} \right\}} \right]

is equal to

View answer and explanation

Correct option: C

Option C (1) is correct because after simplifying the composite function f[f{f(x)}] it results in x, and the derivative of x is 1.

9 Easy

9 of 500

Application +4 | -1

If for

x \in\left(0, \frac{1}{4}\right)

, the derivative of

\tan ^{-1}\left(\frac{6 x \sqrt{x}}{1-9 x^3}\right)

\sqrt{x} \cdot g(x)

, then

g(x)

equals

View answer and explanation

Correct option: B

\frac{9}{1+9 x^3}

Option B (9/(1+9x³)) is correct because after simplifying the inverse tangent expression and differentiating, the resulting g(x) is 9/(1+9x³).

10 Easy

10 of 500

Application +4 | -1

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

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: A

\left(1+\frac{1}{x}\right)^{x}\left[\log \left(1+\frac{1}{x}\right)-\frac{1}{1+x}\right]

Option A (

\left(1+\frac{1}{x}\right)^{x}\left[\log \left(1+\frac{1}{x}\right)-\frac{1}{1+x}\right]

) is correct because it is the result of logarithmic differentiation.

11 Easy

11 of 500

Application +4 | -1

The rate at which fluid level inside vertical cylindrical tank of radius

r

drop if we pump fluid out at the rate of

3 \mathrm{~cm}^3 \mathrm{l} / \mathrm{min}

View answer and explanation

Correct option: B

\frac{3}{\pi r^2}

Option B (

\frac{3}{\pi r^2}

) is correct because it represents the rate of change of height with respect to time when fluid is pumped out at a given rate.

12 Easy

12 of 500

Application +4 | -1

For

\left| x \right| < 1,

let

y = 1 + x + {x^2} + \ldots

\infty ,

then

\frac{{dy}}{{dx}} - y

is equal to

View answer and explanation

Correct option: D

x{y^2}

Option D (

x{y^2}

) is correct because, after finding the derivative

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

and simplifying, the expression

\frac{{dy}}{{dx}} - y

is equal to

x{y^2}

13 Easy

13 of 500

Analyzing +4 | -1

y = 1 - x + \frac{{{x^2}}}{{2!}} - \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} - ...,

then

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

is equal to

View answer and explanation

Correct option: D

y

The correct answer is D (

y

) because the second derivative of the given series is equal to the original series itself.

14 Easy

14 of 500

Application +4 | -1

f(x)=x^2+\tan x

and

f

is inverse of

g

, then

g^{\prime}(x)

is equal to

View answer and explanation

Correct option: D

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

Option D is correct because it uses the inverse function derivative formula and correctly substitutes the derivative of the original function.

15 Easy

15 of 500

Analyzing +4 | -1

f(x-y), f(x) f(y)

and

f(x+y)

are in A.P. for all

x, y \in R

and

f(0)=0

, then

View answer and explanation

Correct option: A

f^{\prime}(-2)=f^{\prime}(2)

Option A (

f^{\prime}(-2)=f^{\prime}(2)

) is correct because the given condition implies that f(x) is an odd function, which means its derivative f'(x) is an even function.

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16 Easy

16 of 500

Analyzing +4 | -1

y = x\log \left( {\frac{x}{{a + bx}}} \right),

then

\frac{{{x^3}{d^2}y}}{{d{x^2}}}

is equal to

View answer and explanation

Correct option: B

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

The correct answer is B (

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

) because it is the result of differentiating the given function and substituting to find the expression for

{x^3}\frac{{{d^2}y}}{{d{x^2}}}

17 Easy

17 of 500

Application +4 | -1

f\left( x \right) = {\log _{{x^2}}}\left( {lo{g_e}x} \right),

then

f'\left( x \right)

x = e

View answer and explanation

Correct option: C

\frac{1}{{2e}}

Option C (

\frac{1}{{2e}}

) is correct because after differentiating and substituting x=e, the result is

\frac{1}{{2e}}

18 Easy

18 of 500

Analyzing +4 | -1

Let the functions

f:R \to R

and

g:R \to R

be defined as

f(x) = \left\{ {\begin{array}{cccccccccccccccccccc} {x + 2,}&{x\,\, < \,\,0}\\ {{x^2},\,}&{x\,\, \ge \,\,0} \end{array}} \right.\,

and

g(x) = \left\{ {\begin{array}{cccccccccccccccccccc} {{x^3},}&{x\,\, < \,\,1}\\ {3x - 2,}&{x\,\, \ge \,\,1} \end{array}} \right.\,

Then, the number of points in

R

where

(fog)\,(x)

is NOT differentiable is equal to :

View answer and explanation

Correct option: A

Option A (1) is correct because the function (fog)(x) is non-differentiable at only one point, x = 0, due to discontinuity.

19 Easy

19 of 500

Understanding +4 | -1

y = {\sin ^{ - 1}}\frac{x}{2} + {\cos ^{ - 1}}\frac{x}{2}

, then the value of

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

View answer and explanation

Correct option: C

Option C (0) is correct because the expression simplifies to a constant, and the derivative of a constant is zero.

20 Easy

20 of 500

Application +4 | -1

f(t)=\frac{t^2-5 t-1}{t^3}

then

f^{\prime}\left(\frac{1}{a}\right)

is given by

View answer and explanation

Correct option: B

3 a^4+10 a^3-a^2

Option B (

3 a^4+10 a^3-a^2

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

t = \frac{1}{a}

21 Easy

21 of 500

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.

22 Easy

22 of 500

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.

23 Easy

23 of 500

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.

24 Easy

24 of 500

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.

25 Easy

25 of 500

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.

26 Easy

26 of 500

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.

27 Easy

27 of 500

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.

28 Easy

28 of 500

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)

29 Easy

29 of 500

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)

30 Easy

30 of 500

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.

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 500 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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