If $ x+|y|=2 y $, then $ y $ as a function of $ x $, at $ x=0 $ is

differentiable but not continuous

continuous as well as differentiable

neither continuous nor differentiable

If $f(x)=\sin ^{-1}\left(3 x-4 x^3\right)$ for all $x \in[-1,1]$ then

$f'(x) = \left\{ {\begin{array}{*{20}{l}} {\frac{3}{{\sqrt {1 - {x^2}} }},}&{{\rm{ if}}\,\,{\rm{ }} - \frac{1}{2} < x < \frac{1}{2}}\\ {\frac{{ - 3}}{{\sqrt {1 - {x^2}} }},}&{{\rm{ if }}\,\,\frac{1}{2} < x < 1\,\,\,{\rm{ or}},\,\,\, - 1 < x < - \frac{1}{

$f^{\prime}(x)=\frac{3}{\sqrt{1-x^2}}$ for all $x \in(-1,1)$

$f^{\prime}(x)=\frac{3}{\sqrt{1-x^2}}$ for all $x \in[-1,1]$

$f^{\prime}(x)= \begin{cases}\frac{3}{\sqrt{1-x^2}}, & \text { if }|x| |x|>\frac{\sqrt{3}}{2}\end{cases}$

If $ y=\frac{2(x-\sin x)^{3 / 2}}{\sqrt{x}} $, then $ \frac{d y}{d x} $ is equal to

$ \frac{2(x-\sin x)^{3 / 2}}{\sqrt{x}}\left[\frac{3}{2} \cdot \frac{1-\cos x}{1-\sin x}-\frac{1}{2 x}\right] $

$ \frac{2(x-\sin x)^{1 / 2}}{\sqrt{x}}\left[\frac{3}{2} \cdot \frac{1-\cos x}{x-\sin x}-\frac{1}{2 x}\right] $

The derivative of ${a^{\sec x}}{\rm{w}}{\rm{.r}}{\rm{.t}}{\rm{.}}{a^{\tan x}}(a > 0)$ is

$\sin x{\rm{ }}{a^{\sec x - \tan x}}$

$\sec x{\rm{ }}{a^{\sec x - \tan x}}$

$\sin x{\rm{ }}{a^{\tan x - \sec x}}$

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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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271 Easy

271 of 500

Analyzing +4 | -1

x+|y|=2 y

, then

y

as a function of

x

, at

x=0

View answer and explanation

Correct option: B

continuous but not differentiable

Option B is correct because the function is continuous at x=0 but not differentiable due to different left-hand and right-hand derivatives.

272 Easy

272 of 500

Application +4 | -1

f(x) = \left\{ {\begin{array}{llllllllllllllllllll} {x - 5\quad \mathrm{ }\,\,\mathrm{for}\,\,\mathrm{ }x \le 1}\\ {4{x^2} - 9\mathrm{ }\,\,\mathrm{for}\,\,\mathrm{ }1 < x < 2,\mathrm{ }\,\mathrm{then }\,\,f'\left( {{2^ + }} \right) = }\\ {3x + 4\mathrm{ }\,\,\mathrm{for}\,\,\mathrm{ }x \ge 2} \end{array}} \right.

View answer and explanation

Correct option: C

The correct answer is C (3) because it is the limit of the derivative of the function as x approaches 2 from the right.

273 Easy

273 of 500

Application +4 | -1

\frac{d}{d x}\left[\cos \left(1-x^{2}\right)^{2}\right]

is equal to

View answer and explanation

Correct option: C

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

The correct answer is C (

4 x(1-x^{2}) \sin (1-x^{2})^{2}

) because it is the result of applying the chain rule multiple times to differentiate the given function.

274 Easy

274 of 500

Analyzing +4 | -1

{y^2} + {\log _e}({\cos ^2}x) = y,\,\,x \in \left( { - \frac{\pi }{2},\,\frac{\pi }{2}} \right),

then

View answer and explanation

Correct option: C

|y''(0)| = 2

Option C (

\left| y''(0) \right| = 2

) is correct because differentiating the given equation and evaluating at x=0 yields |y''(0)| = 2.

275 Easy

275 of 500

Application +4 | -1

Let

f(x)

be a polynomial function of degree 2 such that

f(x)>0

for all

x \in R

. If

g(x)=f(x)+f^{\prime}(x)+f^{\prime \prime}(x)

for all

x

, then

View answer and explanation

Correct option: B

g(x) > 0

for all

x

Option B is correct because it correctly uses the given conditions to determine the sign of the quadratic function g(x).

276 Easy

276 of 500

Application +4 | -1

\frac{d}{d x}\left(\tan ^{-1} \sqrt{\frac{1+\cos \frac{x}{2}}{1-\cos \frac{x}{2}}}\right)

is equal to

View answer and explanation

Correct option: A

-\frac{1}{4}

The correct answer is A (

-\frac{1}{4}

) because after simplifying the expression inside the inverse tangent, the derivative is a constant.

277 Easy

277 of 500

Analyzing +4 | -1

y=\sin ^{-1}(\sin x)

, then

\frac{d y}{d x}

x=\frac{\pi}{2}

, is

View answer and explanation

Correct option: C

Non-existent

Option C (Non-existent) is correct because the left-hand derivative and right-hand derivative of the function at

x = \frac{\pi}{2}

are not equal.

278 Easy

278 of 500

Analyzing +4 | -1

x = {e^{y + {e^y} + ....\infty }};\,\,\,x > 0,

then

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

equals

View answer and explanation

Correct option: C

\frac{{1 - x}}{x}

Option C (

\frac{{1 - x}}{x}

) is correct because it is the derivative of y with respect to x after taking the logarithm and differentiating.

279 Easy

279 of 500

Application +4 | -1

y = {\tan ^{ - 1}}\left[ {\frac{{\sin x + \cos x}}{{\cos x - \sin x}}} \right],

then

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

is equal to

View answer and explanation

Correct option: D

Option D (1) is correct because after simplifying the expression inside the arctangent, the derivative is 1.

280 Easy

280 of 500

Analyzing +4 | -1

f(x)=\sin ^{-1}\left(3 x-4 x^3\right)

for all

x \in[-1,1]

then

View answer and explanation

Correct option: C

f'(x) = \left\{ {\begin{array}{llllllllllllllllllll} {\frac{3}{{\sqrt {1 - {x^2}} }},}&{\mathrm{ if}\,\,\mathrm{ } - \frac{1}{2} < x < \frac{1}{2}}\\ {\frac{{ - 3}}{{\sqrt {1 - {x^2}} }},}&{\mathrm{ if }\,\,\frac{1}{2} < x < 1\,\,\,\mathrm{ or},\,\,\, - 1 < x < - \frac{1}{

Option C is correct because the derivative of arcsin(3x-4x³) is piecewise, being 3/sqrt(1-x²) for |x|<1/2 and -3/sqrt(1-x²) for 1/2<|x|<1 due to the properties of arcsin.

281 Easy

281 of 500

Application +4 | -1

3 f(x)-2 f\left(\frac{1}{x}\right)=x

, then

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

View answer and explanation

Correct option: B

\frac{1}{2}

Option B (

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

) is correct because it is the value of

f^{\prime}(2)

obtained by solving the system of equations after differentiating the given functional equation.

282 Easy

282 of 500

Application +4 | -1

Let

f

be a differentiable function satisfying the condition:

f\left( {\frac{x}{y}} \right) = \frac{{f(x)}}{{f(y)}}\mathrm{ }\,\,\mathrm{for}\,\,\mathrm{ all}\,\,\mathrm{ }x,y( \ne 0) \in R\mathrm{ }\,\,\mathrm{and }\,\,f(y) \ne 0

f^{\prime}(1)=2

, then

f^{\prime}(x)

is equal to

View answer and explanation

Correct option: D

\frac{2 f(x)}{x}

Option D is correct because using the given condition

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

and

f^{\prime}(1)=2

, we find

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

283 Easy

283 of 500

Analyzing +4 | -1

f\left( 4 \right) = 4,f'\left( 4 \right) = 1

, then

\lim\limits_{x \to 4} \frac{{2 - \sqrt {f\left( x \right)} }}{{2 - \sqrt x }}

is equal to

View answer and explanation

Correct option: B

The correct answer is B (1) because applying L'Hopital's rule simplifies the limit to 1.

284 Easy

284 of 500

Application +4 | -1

Let

y(x)=(1+x)\left(1+x^2\right)\left(1+x^4\right) \ldots\left(1+x^{32}\right)

. Then

\frac{d y}{d x}

x=\frac{1}{2}

View answer and explanation

Correct option: C

4-65\left(\frac{1}{2}\right)^{62}

Option D (

4-65\left(\frac{1}{2}\right)^{62}

) is correct because it is the value of the derivative at x=1/2 after simplification.

285 Easy

285 of 500

Analyzing +4 | -1

Let

f(x)

be a second degree polynomial function such that

f(-1)=f(1)

and

\alpha, \beta, \gamma

are in A.P. Then,

f^{\prime}(\alpha), f^{\prime}(\beta), f^{\prime}(\gamma)

are in

View answer and explanation

Correct option: A

Option A (AP) is correct because the derivative of the quadratic function is linear, and since

\alpha, \beta, \gamma

are in AP, their corresponding derivatives will also be in AP.

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286 Easy

286 of 500

Application +4 | -1

\sec \left( {\frac{{x - y}}{{x + y}}} \right) = a

, then

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

View answer and explanation

Correct option: A

\frac{y}{x}

Option A (

\frac{y}{x}

) is correct because it is the correct derivative obtained by implicit differentiation of the given equation.

287 Easy

287 of 500

Application +4 | -1

x^m y^n=(x+y)^{m+n}

, then

\frac{d y}{d x}

View answer and explanation

Correct option: B

\frac{y}{x}

Option B (y/x) is correct because it is the derivative obtained after implicit differentiation and simplification.

288 Easy

288 of 500

Analyzing +4 | -1

The derivative of

f(\tan x)

with respect to

g(\sec x)

x=\frac{\pi}{4}

, where

f^{\prime}(1)=2

and

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

, is

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because after applying the chain rule and simplifying, the derivative evaluates to

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

x = \frac{\pi}{4}

289 Easy

289 of 500

Analyzing +4 | -1

Let

f

be a differentiable function satisfying

f(x)+f(y)+f(z)+f(x) f(y) f(z)=14

for all

x, y, z \in R

Then,

View answer and explanation

Correct option: B

f^{\prime}(x)=0

for all

x \in R

Option D (All the above) is correct because all three given statements are true based on the given functions and their derivatives.

290 Easy

290 of 500

Application +4 | -1

f(x)=|\log x|

, then for

x \neq 1, f^{\prime}(x)

equals

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the derivative of

f(x) = |\log x|

is a piecewise function that depends on the value of x.

291 Easy

291 of 500

Analyzing +4 | -1

The 2nd derivative of

a{\sin ^3}t

with respect to

a{\cos ^3}t

\,t = \frac{\pi }{4}

View answer and explanation

Correct option: A

\frac{{4\sqrt 2 }}{{3a}}

Option A (

\frac{{4\sqrt 2 }}{{3a}}

) is correct because it is the value obtained after calculating the second derivative and substituting

t = \frac{\pi}{4}

292 Easy

292 of 500

Application +4 | -1

\frac{d}{d x}\left[\tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}}\right]

is equal to

View answer and explanation

Correct option: C

\frac{1}{2}

Option C is correct because the expression simplifies to x/2, and its derivative is 1/2.

293 Easy

293 of 500

Analyzing +4 | -1

Let

x = {\log _e}t,t > 0

and

y + 1 = {t^2}.

Then,

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

is equal to

View answer and explanation

Correct option: B

- \frac{1}{2}{e^{ - 4x}}

Option B (

- \frac{1}{2}{e^{ - 4x}}

) is correct because it is the result of differentiating the given equations twice and simplifying.

294 Easy

294 of 500

Application +4 | -1

y=\frac{2(x-\sin x)^{3 / 2}}{\sqrt{x}}

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: B

\frac{2(x-\sin x)^{3 / 2}}{\sqrt{x}}\left[\frac{3}{2} \cdot \frac{1-\cos x}{x-\sin x}-\frac{1}{2 x}\right]

Option B is correct because it follows from logarithmic differentiation and simplification of the derivative of the given function.

295 Easy

295 of 500

Application +4 | -1

Let

f(x)=x[x]

where

[x]

denotes the integral part of

x

. If

x

is not an integer, then

f^{\prime}(x)

is equal to

View answer and explanation

Correct option: A

[x]

Option A (

[x]

) is correct because the derivative of

f(x) = x[x]

when

x

is not an integer is simply

[x]

296 Easy

296 of 500

Analyzing +4 | -1

The derivative of

{a^{\sec x}}\mathrm{w}\mathrm{.r}\mathrm{.t}\mathrm{.}{a^{\tan x}}(a > 0)

View answer and explanation

Correct option: C

\sin x\mathrm{ }{a^{\sec x - \tan x}}

Option C (

\sin x\mathrm{ }{a^{\sec x - \tan x}}

) is correct because after applying the chain rule and simplifying, the derivative of

{a^{\sec x}}

with respect to

{a^{\tan x}}

\sin x\mathrm{ }{a^{\sec x - \tan x}}

297 Easy

297 of 500

Application +4 | -1

x=\sin ^{-1} t

and

y=\log \left(1-t^2\right)

; then

\left.\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\right|_{t=1 / 2}

View answer and explanation

Correct option: A

-8 / 3

Option A (-8/3) is correct because it is the value of the second derivative of

y

with respect to

x

evaluated at

t = 1/2

298 Easy

298 of 500

Application +4 | -1

The function

y=\sin ^{-1} x

satisfies

View answer and explanation

Correct option: B

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

Option B ((1-x²)y'' = xy') is correct because it is the result of differentiating y=sin⁻¹x twice and rearranging the terms.

299 Easy

299 of 500

Analyzing +4 | -1

f(x)

is a polynomial satisfying

f(x)=x^3+x^2 f^{\prime}(1)+x f^{\prime \prime}(2)+f^{\prime \prime \prime}(3)

for all

x \in R.

Then,

View answer and explanation

Correct option: D

All the above

Option D (All the above) is correct because the polynomial f(x) satisfies all the given conditions when the values are substituted.

300 Easy

300 of 500

Application +4 | -1

y

is a function of

x

and

\log \left( {x + y} \right) = 2xy,

the value of

y'\left( 0 \right)

is equal to

View answer and explanation

Correct option: A

Option A (1) is correct because, after differentiating the given equation and substituting x=0, the value of

y'(0)

is found to be 1.

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