If $y = {\log _2}{\log _2}\left( x \right)$, then $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ is equal to

$\frac{{{{\log }_2}e}}{{{{\log }_e}x}}$

$\frac{{{{\log }_2}e}}{{x{{\log }_x}2}}$

$\frac{{{{\log }_2}x}}{{{{\log }_e}2}}$

If $y = {\tan ^{ - 1}}x + {\cot ^{ - 1}}x + {\sec ^{ - 1}} + {\rm{cose}}{{\rm{c}}^{ - 1}}x,$ then $\frac{{dy}}{{dx}}$ is equal to

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

If $y = \log {x^x},$ then the value of $\frac{{dy}}{{dx}}$ is

$\log \left( {\frac{e}{x}} \right)$

$\log \left( {\frac{x}{e}} \right)$

If $y=\sqrt{x+\sqrt{x+\sqrt{x}}}, x > 0$, then $d y / d x$ at $x=1$ is

$\frac{1}{\sqrt{1+\sqrt{2}}}\left(\frac{3+4 \sqrt{2}}{4 \sqrt{2}}\right)$

$\left(\frac{3+4 \sqrt{2}}{8 \sqrt{2}}\right)$

$\frac{3+4 \sqrt{2}}{\sqrt{1+\sqrt{2}}}$

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JEE • MATHEMATICS • DIFFERENTIATION

Differentiation practice · Mixed difficulty

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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 151-180 of 932 questions

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151 Easy

151 of 932

Application +4 | -1

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

, then

d y / d x=

View answer and explanation

Correct option: C

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

Option C (-1/sqrt(1-x²)) is correct because it is the derivative of the given function after simplification using trigonometric substitution.

152 Medium

152 of 932

Application +4 | -1

y = {\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + {x^2}} - 1}}{x}} \right),

then

y'\left( 0 \right)

View answer and explanation

Correct option: A

\frac{1}{2}

Option A (

\frac{1}{2}

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

x=0

after simplifying using trigonometric substitution.

153 Medium

153 of 932

Application +4 | -1

y = {\log _2}{\log _2}\left( x \right)

, then

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

is equal to

View answer and explanation

Correct option: D

\frac{{{{\log }_2}e}}{{x{{\log }_e}x}}

The correct answer is D (

\frac{{{{\log }_2}e}}{{x{{\log }_e}x}}

) because it is the result of correctly applying the chain rule and change of base formula for logarithms.

154 Easy

154 of 932

Application +4 | -1

x=3 \tan t

and

y=3 \sec t

, then the value of

\frac{d^{2} y}{\mathrm{dx}^{2}}

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

, is:

View answer and explanation

Correct option: B

\frac{1}{6 \sqrt{2}}

Option B (

\frac{1}{6 \sqrt{2}}

) is correct because it is the value of the second derivative calculated using parametric differentiation at the given point.

155 Easy

155 of 932

Analyzing +4 | -1

{y^2} = a{x^2} + bx + c

where

a,b,c

are constants, then

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

, is equal to

View answer and explanation

Correct option: A

A constant

Option A (A constant) is correct because after differentiating the given equation twice and simplifying, the expression

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

results in a constant value.

156 Hard

156 of 932

Application +4 | -1

f(x)=|\cos x|

, then

f^{\prime}(3 \pi / 4)

is equal to

View answer and explanation

Correct option: B

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

Option B is correct because it is the result of differentiating the function in the given neighborhood and evaluating at the specified point.

157 Easy

157 of 932

Understanding +4 | -1

y = {\tan ^{ - 1}}x + {\cot ^{ - 1}}x + {\sec ^{ - 1}} + \mathrm{cose}{\mathrm{c}^{ - 1}}x,

then

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

is equal to

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.

158 Easy

158 of 932

Application +4 | -1

y = \log {x^x},

then the value of

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

View answer and explanation

Correct option: B

\log \left( {ex} \right)

Option B (log(ex)) is correct because simplifying y and then differentiating gives log(ex).

159 Medium

159 of 932

Application +4 | -1

f\left( x \right) = \left( {x - 2} \right)\left( {x - 4} \right)\left( {x - 6} \right) \ldots \left( {x - 2n} \right),

then

f'\left( 2 \right)

View answer and explanation

Correct option: B

{\left( { - 2} \right)^{n - 1}}\left( {n - 1} \right)!

Option B (

{(- 2)^{n - 1}}(n - 1)!

) is correct because after differentiating the given function and evaluating at x=2, the result is

{(- 2)^{n - 1}}(n - 1)!

160 Medium

160 of 932

Application +4 | -1

f(x)=\sqrt{1+\cos ^{2}\left(x^{2}\right)}

, then

f^{\prime}\left(\frac{\sqrt{\pi}}{2}\right)

View answer and explanation

Correct option: B

-\sqrt{(\pi / 6)}

The correct answer is B (

-\sqrt{(\pi / 6)}

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

x=\frac{\sqrt{\pi}}{2}

, using the chain rule and the derivatives of trigonometric functions.

161 Easy

161 of 932

Analyzing +4 | -1

The functions

u=e^x \sin x ; v=e^x \cos x

satisfy the equation

View answer and explanation

Correct option: D

All the above

Option D (All the above) is correct because the given functions satisfy all the equations provided when their derivatives are calculated and substituted.

162 Hard

162 of 932

Analyzing +4 | -1

Let

f(x) = 15 - \left| {x - 10} \right|;\,\,x \in R

. Then, the set of all values of

x

, at which the function,

g(x) = f(f(x))

is not differentiable, is

View answer and explanation

Correct option: B

\left\{ {5,\,10,\,15} \right\}

Option B ({5, 10, 15}) is correct because these are the points where the absolute value function inside g(x) changes its behavior, leading to non-differentiability.

163 Easy

163 of 932

Application +4 | -1

The derivative of the function

2 x^2+6 x+1

x=3

View answer and explanation

Correct option: A

Option A (18) is correct because substituting x=3 into the derivative of the function yields 18.

164 Easy

164 of 932

Application +4 | -1

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

and

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

then

d y / d x

x=1

View answer and explanation

Correct option: D

\frac{1}{2} \sin ^2\left(\frac{1}{2}\right)

Option D (

\frac{1}{2} \sin ^2(\frac{1}{2})

) is correct because, after applying the chain rule and simplifying, we evaluate the derivative at

x=1

, which gives

\frac{1}{2} \sin ^2(\frac{1}{2})

165 Medium

165 of 932

Application +4 | -1

f:(0, \infty) \rightarrow \mathbb{R}

is continuous. If

F(x)

is a differentiable function such that

F^{\prime}(x)=f(x) \forall x>0

and

F\left(x^2\right)=x^2+x^3

, then

f(4)

equals

View answer and explanation

Correct option: C

Option C (4) is correct because by differentiating F(x²) and using F'(x) = f(x), we can solve for f(4).

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166 Medium

166 of 932

Application +4 | -1

\sin y + {e^{ - x\cos y}} = e

, then

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

\left( {1,\pi } \right)

View answer and explanation

Correct option: C

e

Option D (

e

) is correct because after differentiating the given equation implicitly and substituting x=1 and y=

\pi

, the value of

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

is found to be e.

167 Easy

167 of 932

Analyzing +4 | -1

The second order derivative of

a{\sin ^3}a{\sin ^3}t

with respect to

a{\cos ^3}{\cos ^3}t\,

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

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because after finding the second-order derivative and evaluating it at

t = \frac{\pi }{4}

, the expression simplifies to the stated result.

168 Easy

168 of 932

Application +4 | -1

2f\left( x \right) = f'\left( x \right)

and

f\left( 0 \right) = 3,

then

f\left( 2 \right)

is equal to

View answer and explanation

Correct option: A

3{e^4}

Option A (

3{e^4}

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

169 Easy

169 of 932

Application +4 | -1

y=x^{x^{x^{x^{x^{...}}}}}

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: C

\frac{y}{x(1+y \log x)}

Option B is correct because it is derived by taking the logarithm of both sides and then differentiating implicitly.

170 Medium

170 of 932

Analyzing +4 | -1

Let

y = {t^{10}} + 1

and

x = {t^8} + 1,

then

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

is equal to

View answer and explanation

Correct option: C

\frac{5}{{16{t^6}}}

Option C (

\frac{5}{{16{t^6}}}

) is correct because after finding the first and second derivatives using parametric differentiation, the expression simplifies to this result.

171 Medium

171 of 932

Application +4 | -1

{10^{ - x\tan x}}\left[ {\frac{d}{{dx}}\left( {{{10}^{x\tan x}}} \right)} \right]

is equal to

View answer and explanation

Correct option: B

In 10

(\tan x + x{\sec ^2}x)

Option B (In 10

(\tan x + x{\sec ^2}x)

) is correct because it is the result of differentiating

10^{x\tan x}

and then simplifying.

172 Medium

172 of 932

Analyzing +4 | -1

Let

y(x)=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)\left(1+x^{8}\right)\left(1+x^{16}\right)

. Then

y^{\prime}-y^{\prime \prime}

x=-1

is equal to :

View answer and explanation

Correct option: A

496

Option A (496) is correct because after simplifying the expression for y(x), we can find the first and second derivatives and evaluate them at x = -1, then compute y' - y''.

173 Easy

173 of 932

Application +4 | -1

y=\sqrt{x+\sqrt{x+\sqrt{x}}}, x > 0

, then

d y / d x

x=1

View answer and explanation

Correct option: A

\frac{1}{2 \sqrt{1+\sqrt{2}}}\left(\frac{3+4 \sqrt{2}}{4 \sqrt{2}}\right)

Option A is correct because it is the result of applying the chain rule multiple times and evaluating the derivative at x=1.

174 Easy

174 of 932

Application +4 | -1

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

, then

\frac{d y}{d x}

is equal to :

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because it is the derivative of arccot(x²) using the chain rule.

175 Medium

175 of 932

Application +4 | -1

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

, then

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

is equal to

View answer and explanation

Correct option: B

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

Option B (

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

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

f^{-1}(x)

after simplifying

f(x)

and finding its inverse.

176 Medium

176 of 932

Understanding +4 | -1

f(x)=\left(\frac{x^a}{x^b}\right)^{a+b} \cdot\left(\frac{x^b}{x^c}\right)^{b+c} \cdot\left(\frac{x^c}{x^a}\right)^{c+a}

then

f^{\prime}(x)

is equal to

View answer and explanation

Correct option: B

Option B (0) is correct because after simplification, f(x) becomes 1, and the derivative of a constant is 0.

177 Medium

177 of 932

Analyzing +4 | -1

Let

f:R \to R

be a function such that

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

Then

f(2)

equals

View answer and explanation

Correct option: C

-2

Option C (-2) is correct because by finding the derivatives and solving the resulting system of equations, we can determine f(x) and then evaluate f(2).

178 Medium

178 of 932

Application +4 | -1

f(x)

and

g(x)

are two functions from

\mathbb{R}

\mathbb{R}

such that

(f \circ g)(x)=\left(x^3-x^2+2\right)^8

, then

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

View answer and explanation

Correct option: A

Option A (8) is correct because, by choosing

f(x) = x^8

and

g(x) = x^3 - x^2 + 2

, we get

(f \circ g)(x) = (x^3 - x^2 + 2)^8

. Then,

f'(1) = 8

and

g'(1) = 1

, so

f'(1)g'(1) = 8

179 Medium

179 of 932

Application +4 | -1

Let

f(x)=\frac{x^{2}-x}{x^{2}+2 x} x \neq 0,-2

. Then

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

(wherever it is defined) is equal to :

View answer and explanation

Correct option: B

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

Option B (

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

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

180 Medium

180 of 932

Application +4 | -1

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

, then

f^{\prime}\left(\frac{\sqrt{\pi}}{2}\right)

View answer and explanation

Correct option: B

-\sqrt{\frac{\pi}{6}}

Option B (

-\sqrt{\frac{\pi}{6}}

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

x = \frac{\sqrt{\pi}}{2}

Revision notes and key takeaways

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