If $f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{{x\log \cos x}}{{\log \left( {1 + {x^2}} \right)}},x \ne 0}\\ {0,x = 0} \end{array}} \right.$, then

$f\left( x \right)$ is not continuous at $x = 0$

$f\left( x \right)$ is not continuous at $x = 0$ but not differentiable at $x = 0$

Let $f: R \rightarrow R$ be a function such that $f(x + y) = f(x) + f(y){\rm{ }}\,{\rm{for}}\,{\rm{ all }}\,x,y \in R$ If $f(x)$ is differentiable at $x=0$, then which one of the following is incorrect?

$f(x)$ is continuous for all $x \in R$

$f^{\prime}(x)$ is constant for all $x \in R$

If $f(x)=\cos ^{-1}\left(2 x^2-1\right), x \in[-1,1]$. Then,

$f(x)$ is differentiable on $(-1,0) \cup(0,1)$ such that $f'(x) = \left\{ {\begin{array}{*{20}{l}} {\frac{{ - 2}}{{\sqrt {1 - {x^2}} }},}&{0 < x < 1}\\ {\frac{2}{{\sqrt {1 - {x^2}} }},}&{ - 1 < x < 0} \end{array}} \right.$

$f(x)$ is differentiable on $(-1,1)$ such that $f^{\prime}(x)=\frac{-2}{\sqrt{1-x^2}}$

$f(x)$ is differentiable on $(-1,0) \cup(0,1)$ such that $f'(x) = \frac{{ - 2}}{{\sqrt {1 - {x^2}} }}$

$f(x)$ is differentiable on $(-1,1)$ such that $f'(x) = \left\{ {\begin{array}{*{20}{l}} {\frac{{ - 2}}{{\sqrt {1 - {x^2}} }},}&{0 \le x < 1}\\ {\frac{2}{{\sqrt {1 - {x^2}} }},}&{ - 1 < x \le 0} \end{array}} \right.$

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

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

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

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

If $y=A \sin \omega t$ then $\frac{d^5 y}{d t^5}=$

${A} \omega^5 \cos \left(\omega {t}-\frac{\pi}{2}\right)$

${A} \omega^5 \cos \left(\omega t+\frac{\pi}{2}\right)$

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JEE • MATHEMATICS • DIFFERENTIATION

Differentiation practice · Easy

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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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241 Easy

241 of 500

Understanding +4 | -1

The number of inflection points of a function given by a third degree polynomial is exactly

View answer and explanation

Correct option: B

Option B (1) is correct because a third-degree polynomial has exactly one inflection point.

242 Easy

242 of 500

Application +4 | -1

f(1) = 1,\,\,f'(1) = 3,

then the derivative of

f(f(f(x))) + {(f(x))^2}\,\,\mathrm{at }\,\,x = 1

View answer and explanation

Correct option: D

Option D (33) is correct because it is the result of applying the chain rule and the given values of

f(1)

and

f'(1)

to the derivative of the composite function.

243 Easy

243 of 500

Analyzing +4 | -1

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{x\log \cos x}}{{\log \left( {1 + {x^2}} \right)}},x \ne 0}\\ {0,x = 0} \end{array}} \right.

, then

View answer and explanation

Correct option: B

f\left( x \right)

is not continuous and differentiable at

x = 0

Option D (None of these) is incorrect. The function is continuous and differentiable at x=0. The solution steps show the derivative exists and is -1/2.

244 Easy

244 of 500

Analyzing +4 | -1

Let

f: R \rightarrow R

be a function such that

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

f(x)

is differentiable at

x=0

, then which one of the following is incorrect?

View answer and explanation

Correct option: D

f(x)

is differentiable only in a finite interval containing zero

Option D (

f(x)

is differentiable only in a finite interval containing zero) is correct because the given condition implies that

f(x)

is differentiable for all

x \in R

, and

f'(x)

is constant for all

x \in R

245 Easy

245 of 500

Application +4 | -1

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

equals

View answer and explanation

Correct option: B

\frac{1}{2}

Option B (

\frac{1}{2}

) is correct because after simplifying the expression using trigonometric identities, the derivative is simply

\frac{1}{2}

246 Easy

246 of 500

Analyzing +4 | -1

(a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^{2}-b^{2}

, where

a>b>0

, then

\frac{d x}{d y}

\left(\frac{\pi}{4}, \frac{\pi}{4}\right)

is:

View answer and explanation

Correct option: C

\frac{a+b}{a-b}

Option C is correct because by differentiating the given equation implicitly and evaluating at the specified point, we find that dx/dy = (a+b)/(a-b).

247 Easy

247 of 500

Understanding +4 | -1

Let

f

be a function defined for every

x

, such that

f^{\prime \prime}

=-f, f(0)=0, f^{\prime}(0)=1

then

f(x)

is equal to

View answer and explanation

Correct option: C

\sin x

Option C (

\sin x

) is the correct solution because it satisfies the given differential equation and initial conditions.

248 Easy

248 of 500

Analyzing +4 | -1

\sqrt {1 - {x^6}} + \sqrt {1 - {y^6}} = a\left( {{x^3} - {y^3}} \right)

and

\frac{{dy}}{{dx}} = f\left( {x,y} \right)\sqrt {\frac{{1 - {y^6}}}{{1 - {x^6}}}}

, then

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

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because by substituting

{x^3} = \sin A

and

{y^3} = \sin B

, differentiating, and simplifying, we find that

f(x,y) = \frac{{{x^2}}}{{{y^2}}}

249 Easy

249 of 500

Analyzing +4 | -1

Let

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {1,\forall x < 0}\\ {1 + \sin x,\forall 0 \le x \le \pi /2} \end{array}} \right.

, then what is the value of

f'\left( x \right)

x = 0

View answer and explanation

Correct option: D

Does not exist

Option D (Does not exist) is correct because the left-hand derivative (LHD) and right-hand derivative (RHD) at x=0 are not equal.

250 Easy

250 of 500

Application +4 | -1

Let

f_1(x)=e^x

and

f_{n+1}(x)=e^{f_n(x)}

for any

n \geq 1, n \in N

. Then for any fixed

n

, the value of

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

equals :

View answer and explanation

Correct option: B

f_n(x) f_{n-1}(x) \ldots \ldots f_2(x) f_1(x)

Option B (

f_n(x) f_{n-1}(x) \ldots f_2(x) f_1(x)

) is correct because it is the result of repeatedly applying the chain rule.

251 Easy

251 of 500

Application +4 | -1

y = {\tan ^{ - 1}}\left( {\frac{{a\cos x - b\sin x}}{{b\cos x + a\sin x}}} \right),

then

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

is equal to

View answer and explanation

Correct option: B

- 1

Option B (-1) is correct because by simplifying the expression inside the inverse tangent function and differentiating, we get a constant derivative of -1.

252 Easy

252 of 500

Analyzing +4 | -1

Let

f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {\frac{{1 - x}}{x}} } \right)} \right),\,\,0 < x < 1

. Then,

View answer and explanation

Correct option: C

{(1 - x)^2}f'(x) + 2{(f(x))^2} = 0

The correct answer is C (

{(1 - x)^2}f'(x) + 2{(f(x))^2} = 0

) because it is the relationship derived after simplifying the function and finding its derivative.

253 Easy

253 of 500

Application +4 | -1

y=f\left(x^3\right), z=g\left(x^2\right), f^{\prime}(x)=\cos x

and

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

, then

\frac{d y}{d z}

View answer and explanation

Correct option: A

\frac{3 x}{2} \cos x^3 \operatorname{cosec} x^2

Option A is correct because it follows from applying the chain rule to find dy/dx and dz/dx and then dividing them.

254 Easy

254 of 500

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

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

Option D (

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

) is correct because it is the result of applying the chain rule.

255 Easy

255 of 500

Application +4 | -1

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

, then

f^{\prime}(x)

x=e

View answer and explanation

Correct option: B

\frac{1}{e}

The correct answer is B (

\frac{1}{e}

) because it is the value of the derivative of the given function evaluated at x=e, using the quotient rule and the change of base formula for logarithms.

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256 Easy

256 of 500

Application +4 | -1

e^{f(x)}=\log _e x

and

g(x)

is the inverse function of

f(x)

, then

g^{\prime}(x)

is equal to :

View answer and explanation

Correct option: C

e^{e^x+x}

Option D (

e^{e^x+x}

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

g(x)

after finding

f(x)

and

g(x)

257 Easy

257 of 500

Analyzing +4 | -1

The derivative of

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

with respect to

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

, is

View answer and explanation

Correct option: C

1 for

|x|<1

and

-1

for

|x|>1

Option C (1 for |x|<1 and -1 for |x|>1) is correct because by using the substitution

x = tan(\theta)

, we can simplify both inverse trigonometric functions and then find the derivative.

258 Easy

258 of 500

Application +4 | -1

\frac{d y}{d x}=\cos (x+y), y\left(\frac{\pi}{2}\right)=0

, then

y(0)=

View answer and explanation

Correct option: D

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

Option D (

-2 \tan ^{-1}(\frac{\pi}{2}-1)

) is correct because solving the differential equation with the given initial condition and then evaluating at x=0 gives this result.

259 Easy

259 of 500

Application +4 | -1

f(x)=\tan ^{-1}\left(\frac{\sin x}{1+\cos x}\right)

, then

f^{\prime}\left(\frac{\pi}{3}\right)

is equal to

View answer and explanation

Correct option: B

\frac{1}{2}

Option B is correct because the function simplifies to x/2, and its derivative is 1/2, which is constant.

260 Easy

260 of 500

Analyzing +4 | -1

f(x)=\cos ^{-1}\left(2 x^2-1\right), x \in[-1,1]

. Then,

View answer and explanation

Correct option: C

f(x)

is differentiable on

(-1,0) \cup(0,1)

such that

f'(x) = \left\{ {\begin{array}{llllllllllllllllllll} {\frac{{ - 2}}{{\sqrt {1 - {x^2}} }},}&{0 < x < 1}\\ {\frac{2}{{\sqrt {1 - {x^2}} }},}&{ - 1 < x < 0} \end{array}} \right.

Option C is correct because the derivative of arccos(2x^2-1) is piecewise, being -2/sqrt(1-x²) for 0<x<1 and 2/sqrt(1-x²) for -1<x<0 due to the properties of arccos.

261 Easy

261 of 500

Application +4 | -1

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

equals

View answer and explanation

Correct option: D

- \left( {\frac{{{d^2}y}}{{d{x^2}}}} \right){\left( {\frac{{dy}}{{dx}}} \right)^{ - 3}}

Option D (

- \left( {\frac{{d^2y}}{{dx^2}}} \right){\left( {\frac{{dy}}{{dx}}} \right)^{ - 3}}

) is correct because it represents the correct application of the chain rule and inverse function differentiation.

262 Easy

262 of 500

Application +4 | -1

y \sqrt{1-x^2}=k-x \sqrt{1-y^2}

and

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

, then

\frac{d y}{d x}

x=\frac{1}{2}

View answer and explanation

Correct option: C

-\frac{\sqrt{5}}{2}

Option C (-

\frac{\sqrt{5}}{2}

) is correct because it is the value of the derivative of y with respect to x at the specified point.

263 Easy

263 of 500

Application +4 | -1

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

then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: C

2^{x-y} \cdot \frac{2^y-1}{1-2^x}

Option C (

2^{{x-y}} \cdot \frac{{2^y-1}}{{1-2^x}}

) is correct because it is derived by differentiating the given equation implicitly and solving for

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

264 Easy

264 of 500

Understanding +4 | -1

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

, then

\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.

265 Easy

265 of 500

Application +4 | -1

{x^y} = {e^{x - y}},x > 0

then the value of

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

at (1,1) is

View answer and explanation

Correct option: A

Option A (0) is correct because after taking the logarithm and differentiating, the derivative

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

evaluated at (1,1) is 0.

266 Easy

266 of 500

Application +4 | -1

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

then

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

is equal to

View answer and explanation

Correct option: B

- \frac{x}{y}

Option B (

\frac{{ - x}}{y}

) is correct because it is the derivative dy/dx obtained by differentiating the given equation and solving for dy/dx.

267 Easy

267 of 500

Application +4 | -1

y=A \sin \omega t

then

\frac{d^5 y}{d t^5}=

View answer and explanation

Correct option: D

{A} \omega^5 \sin \left(\omega {t}+\frac{\pi}{2}\right)

Option D (

{A} \omega^5 \sin \left(\omega {t}+\frac{\pi}{2}\right)

) is correct because it represents the fifth derivative of the given function

y=A \sin \omega t

, obtained by repeatedly differentiating and using trigonometric identities.

268 Easy

268 of 500

Analyzing +4 | -1

f(x)=\left\{\begin{aligned} e^x ; & x \leq 0 \\ |1-x| ; & x>0\end{aligned}\right.

, then

View answer and explanation

Correct option: B

f(x)

is continuous at

\mathrm{x}=0

Option B (

f(x)

is continuous at

x=0

) is correct because the left and right limits are equal to the function's value at x=0.

269 Easy

269 of 500

Application +4 | -1

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

and

y = {\cos ^{ - 1}}\left( {\sqrt {1 - {t^2}} } \right)

, then

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

is equal to

View answer and explanation

Correct option: D

\frac{1}{3}

Option D (

\frac{1}{3}

) is correct because after substituting

t = \sin \theta

, the derivatives can be easily calculated and the ratio gives the answer.

270 Easy

270 of 500

Application +4 | -1

f(x)=x+\frac{x^2}{1 !}+\frac{x^3}{2 !}+\ldots+\frac{x^n}{(n-1) !}

, then

f(0)+f^{\prime}(0)+f^{\prime \prime}(0)+\ldots+f^n(0)

is equal to

View answer and explanation

Correct option: A

\frac{n(n+1)}{2}

Option A (

\frac{n(n+1)}{2}

) is correct because the sum of the derivatives evaluated at 0 forms an arithmetic progression.

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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