If $y=\sin x+e^x$, then $\frac{d^2 x}{d y^2}$ equals

$\left(-\sin x+e^x\right)^{-1}$

$\frac{\sin x-e^x}{\left(\cos x+e^x\right)^2}$

For $x \in \left( {0,\frac{1}{4}} \right),$ if the derivative of ${\tan ^{ - 1}}\left( {\frac{{6x\sqrt x }}{{1 - 9{x^3}}}} \right)$ is $\sqrt x \cdot g\left( x \right)$, then $g(x)$ equals

$\frac{{3x\sqrt x }}{{1 - 9{x^3}}}$

If $ y=\sqrt{\frac{1+e^{x}}{1-e^{x}}} $, then $ \frac{d y}{d x} $ is equal to

$ \frac{e^{x}}{\left(1-e^{x}\right) \sqrt{1-e^{x}}} $

$ \frac{e^{x}}{\left(1-e^{x}\right) \sqrt{1+e^{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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Showing 451-480 of 500 questions

451 Easy

451 of 500

Analyzing +4 | -1

Let

f(x)=[|x|]

where [.] denotes the greatest integer function, then

f^{\prime}(-1)

View answer and explanation

Correct option: C

non-existent

Option C (non-existent) is correct because the left-hand derivative and right-hand derivative at x=-1 are not equal, making the function non-differentiable.

452 Easy

452 of 500

Application +4 | -1

Let

\mathrm{F(x)=f(x) g(x) h(x)}

for all real

\mathrm{x}

, where

\mathrm{f(x), g(x)}

and

\mathrm{h}(\mathrm{x})

are differentiable functions. At some point

\mathrm{x}_0, \mathrm{F}^{\prime}\left(\mathrm{x}_0\right)=21 \mathrm{F}\left(\mathrm{x}_0\right), \mathrm{f}^{\prime}\left(\mathrm{x}_0\right)=4 \mathrm{f}\left(\mathrm{x}_0\right), \mathrm{g}^{\prime}\left(\mathrm{x}_0\right)=-7 \mathrm{g}\left(\mathrm{x}_0\right)

and

\mathrm{h}^{\prime}\left(\mathrm{x}_0\right)=\operatorname{kh}\left(\mathrm{x}_0\right)

. Then the value of

\mathrm{k} / 8

View answer and explanation

Correct option: A

Option A (3) is correct because by differentiating F(x) and substituting the given conditions, the value of k/8 is found to be 3.

453 Easy

453 of 500

Application +4 | -1

y=\cos ^{-1}\left(\frac{4 \cos x-5 \sin x}{\sqrt{41}}\right)

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: B

Option B (1) is correct because the function simplifies to y = x + constant, making the derivative equal to 1.

454 Easy

454 of 500

Application +4 | -1

\sin ^{-1}[\sqrt{x-a x}-\sqrt{a-a x}]

, then

\frac{d y}{d x}=

View answer and explanation

Correct option: C

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

Option C (

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

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

455 Easy

455 of 500

Analyzing +4 | -1

y=\sin x+e^x

, then

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

equals

View answer and explanation

Correct option: C

\frac{\sin x-e^x}{\left(\cos x+e^x\right)^3}

Option C is correct because it involves finding the second derivative of x with respect to y using the chain rule and inverse function differentiation.

456 Easy

456 of 500

Application +4 | -1

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

, then

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

is equal to

View answer and explanation

Correct option: B

-e

Option B is correct because it correctly simplifies the inverse cosine function and finds its derivative, then evaluates it at the given point.

457 Easy

457 of 500

Application +4 | -1

f(x)=\frac{x^2-x}{x^2+2 x}

, then

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

is equal to

View answer and explanation

Correct option: B

\frac{3}{(1-x)^2}

Option B is correct because it correctly finds the inverse function and then differentiates it.

458 Easy

458 of 500

Analyzing +4 | -1

Derivative of

{\sec ^{ - 1}}\left( {\frac{1}{{1 - 2{x^2}}}} \right)\mathrm{w}\mathrm{.r}\mathrm{.t}\mathrm{.}{\sin ^{ - 1}}\left( {3x - 4{x^3}} \right)

View answer and explanation

Correct option: D

\frac{2}{3}

Option D (2/3) is correct because by substituting

x = \sin \theta

, the expression simplifies, and the derivative can be easily found.

459 Easy

459 of 500

Application +4 | -1

f(x)=x^n

, then the value of

f(1)+\frac{f^{\prime}(1)}{1 !}+\frac{f^{\prime \prime}(1)}{2 !}+\frac{f^{\prime \prime \prime}(1)}{3 !}+\ldots .+\frac{f^{(n)}(1)}{n !}

, is

View answer and explanation

Correct option: B

2^n

Option B is correct because it represents the sum of the binomial coefficients, which equals 2^n.

460 Easy

460 of 500

Analyzing +4 | -1

f\left( x \right) = {\cos ^2}x + {\cos ^2}\left( {x + \frac{\pi }{3}} \right) + \sin x\sin \left( {x + \frac{\pi }{3}} \right)

and

g\left( {\frac{5}{4}} \right) = 3

, then

\frac{d}{{dx}}\left( {gof\left( x \right)} \right) =

View answer and explanation

Correct option: B

Option B (0) is correct because f(x) simplifies to a constant,

\frac{5}{4}

, making gof(x) also a constant, and the derivative of a constant is 0.

461 Easy

461 of 500

Analyzing +4 | -1

Let

f(x y)=f(x) f(y)

for all

x>0, y>0

and

f(1+x)=1+x\{1+g(x)\}

, where

\quad \lim\limits _{x \rightarrow 0} g(x)=0

, then

\int \frac{f(x)}{f^{\prime}(x)} d x

, is

View answer and explanation

Correct option: A

\frac{x^2}{2}+C

Option A (

\frac{x^2}{2}+C

) is correct because after finding

f(x) = x

, the integral

\int \frac{f(x)}{f^{\prime}(x)} dx

simplifies to

\int x dx

, which equals

\frac{x^2}{2}+C

462 Easy

462 of 500

Application +4 | -1

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

, then

d y / d x=

View answer and explanation

Correct option: D

Option D (1) is correct because simplifying the expression inside the arctangent yields

\tan(\frac{{\pi}}{{4}} + x)

, so

y = \frac{{\pi}}{{4}} + x

, and

\frac{{dy}}{{dx}} = 1

463 Easy

463 of 500

Application +4 | -1

x=a\left(\cos t+\log \tan \frac{t}{2}\right), y=a \sin t

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: A

\tan t

Option A (

\tan t

) is correct because it is the simplified form of

\frac{dy}{dx}

after applying trigonometric identities.

464 Easy

464 of 500

Application +4 | -1

For

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

if the derivative of

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

\sqrt x \cdot g\left( x \right)

, then

g(x)

equals

View answer and explanation

Correct option: A

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

Option D (

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

) is incorrect. Option A (

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

) is correct because it is the derivative of the given function after simplification using the arctangent double angle formula.

465 Easy

465 of 500

Analyzing +4 | -1

y = x + {x^2} + {x^3} \ldots \infty ,

where

\left| x \right| < 1,

then for

\left| y \right| < 1,\frac{{dx}}{{dy}}

is equal to

View answer and explanation

Correct option: C

1 - 2y + 3{y^2} - ...\infty

The correct answer is C (

1 - 2y + 3{y^2} - ...\infty

) because it is derived by expressing x in terms of y and then differentiating the resulting series.

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466 Easy

466 of 500

Analyzing +4 | -1

f(x)=\frac{1}{1-x}

, then the derivative of the composite function

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

is equal to

View answer and explanation

Correct option: C

Option C (1) is correct because the composite function simplifies to f[f{f(x)}] = x, and the derivative of x is 1.

467 Easy

467 of 500

Application +4 | -1

y = \frac{{a + b{x^{3/2}}}}{{{x^{5/4}}}}

and

y' = 0

x = 5

, then the ratio

a:b

is equal to

View answer and explanation

Correct option: A

\sqrt 5 :1

Option A (√5 : 1) is correct because setting the derivative to zero at x=5 and solving for the ratio a:b gives √5 : 1.

468 Easy

468 of 500

Application +4 | -1

x=e^{y+e+\cdots x} x>0

, then

\frac{d y}{d x}

View answer and explanation

Correct option: A

\frac{1-x}{x}

Option A ((1-x)/x) is correct because it's the derivative dy/dx obtained after logarithmic differentiation.

469 Easy

469 of 500

Analyzing +4 | -1

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

then

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

is equal to

View answer and explanation

Correct option: C

y

Option C (

y

) is correct because the given series is the Taylor series expansion of

e^{-x}

, and the second derivative of

e^{-x}

e^{-x}

, which is equal to

y

470 Easy

470 of 500

Analyzing +4 | -1

{x^y} = {y^x},

then

x(x - y\log x)\frac{{dy}}{{dx}}

is equal to

View answer and explanation

Correct option: A

y(y - x\log y)

Option A (

y(y - x\log y)

) is correct because after differentiating the given equation and rearranging, the expression simplifies to the stated result.

471 Easy

471 of 500

Application +4 | -1

A man who is

1.5 \mathrm{~m}

tall walks away from a lamp which is

3 \mathrm{~m}

above the ground at the rate of

30 \mathrm{~m} / \mathrm{min}

the shadow lengthening at

\mathrm{K} \mathrm{m} / \mathrm{min}

. Then

\mathrm{K}=

View answer and explanation

Correct option: D

Option D (30) is correct because the rate at which the shadow lengthens is equal to the rate at which the man walks away from the lamp.

472 Easy

472 of 500

Analyzing +4 | -1

Observe the following statements:

I. If

f\left( x \right) = a{x^{41}} + b{x^{ - 40}}

, then

\frac{{f''\left( x \right)}}{{f\left( x \right)}} = 1640{x^{ - 2}}

II.

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

Which of the following is correct?

View answer and explanation

Correct option: A

I is true, but II is false

Option A (I is true, but II is false) is correct because statement I is verified to be true through differentiation, while statement II is false due to the range of the inverse tangent function.

473 Easy

473 of 500

Analyzing +4 | -1

f(x)=\frac{a+\sqrt{a^2-x^2}+x}{\sqrt{a^2-x^2}+a-x}

, where

a>0

and

x < a

, then

f'(0)

has the value equal to

View answer and explanation

Correct option: D

\frac{1}{a}

Option D (

\frac{1}{a}

) is correct because after simplifying the function and finding its derivative, the value of

f'(0)

\frac{1}{a}

474 Easy

474 of 500

Application +4 | -1

The derivative of

{\cos ^3}x

w.r.t.

{\sin ^3}x

View answer and explanation

Correct option: A

- \cot x

Option A (

- \cot x

) is correct because it is the result of dividing the derivative of

{\cos ^3}x

by the derivative of

{\sin ^3}x

475 Easy

475 of 500

Application +4 | -1

y=\left(x^2+1\right)^{\sin x}

, then

y^{\prime}(0)

is equal to

View answer and explanation

Correct option: C

Option C (0) is correct because differentiating

y

and substituting

x=0

results in

y'(0) = 0

476 Easy

476 of 500

Application +4 | -1

f(x)=|\cos x-\sin x|

, then

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

is equal to

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the left-hand derivative and right-hand derivative at x = pi/4 are not equal, so the derivative does not exist.

477 Easy

477 of 500

Analyzing +4 | -1

The derivative of the function,

f(x) = {\cos ^{ - 1}}\left\{ {\frac{1}{{\sqrt {13} }}} \right.(2\cos x - 3\sin x)\} \mathrm{ } + {\sin ^{ - 1}}\left\{ {\frac{1}{{\sqrt {13} }}(2\cos x + 3\sin x)} \right\}

\mathrm{ w}\mathrm{.r}\mathrm{.t}\mathrm{.}\,\,\,\sqrt {1 + {x^2}} \,\,\,\mathrm{at}\,\,\,x = \frac{3}{4}\,\,\mathrm{is}

View answer and explanation

Correct option: C

\frac{10}{3}

Option C (

\frac{10}{3}

) is correct because after simplification and differentiation, the value of the derivative at

x = \frac{3}{4}

\frac{10}{3}

478 Easy

478 of 500

Application +4 | -1

y=\sqrt{\frac{1+e^{x}}{1-e^{x}}}

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: A

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

Option A is correct because the derivative is found using the chain rule and quotient rule, simplifying to the given expression.

479 Easy

479 of 500

Application +4 | -1

Let

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

and

y=f\left(x^2+1\right)

then

\frac{d y}{d x}

x=1

is :

View answer and explanation

Correct option: C

2 \sin 4

Option C (

2 \sin 4

) is correct because it is the result of applying the chain rule and evaluating the derivative at x=1.

480 Easy

480 of 500

Analyzing +4 | -1

Let

f(x)=\left\{\begin{array}{cc}(x-1) \sin \frac{1}{x-1} & \text { if } x \neq 1 \\ 0 & \text { if } x=1\end{array}\right.

Then which one of the following is true?

View answer and explanation

Correct option: B

f

is differentiable at

x=0

but not at

x=1

Option B is correct. The function is differentiable at x=0 because the derivative exists at that point, but it is not differentiable at x=1 because the limit does not exist.

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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