Let $f\left( x \right) = \sin x,{\rm{g}}\left( x \right) = {x^2}$ and $h\left( x \right) = {\log _e}x.$ If $F\left( x \right) = \left( {ho{\rm{g}}\,\,of} \right)\left( x \right),$ then $F''\left( x \right)$ is equal to

$ - 2\,{\rm{cose}}{{\rm{c}}^2}x$

$a\,\,{\rm{cose}}{{\rm{c}}^3}x$

$2\cot {x^2} - 4{x^2}{\rm{cose}}{{\rm{c}}^2}{x^2}$

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

$\frac{\phi^{\prime \prime}}{\psi^{\prime \prime}}$

$\frac{\psi^{\prime \prime}}{\phi^{\prime \prime}}$

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

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

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

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

$f^{\prime}(x)=\left\{\begin{array}{l}\frac{-3}{\sqrt{1-x^2}}, \text { if }|x|<\frac{1}{2} \\ \frac{3}{\sqrt{1-x^2}}, \text { if } \frac{1}{2}<|x|<1\end{array}\right.$

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}} $

If $f(x)=|x|^{|\sin x|}$, then $f^{\prime}\left(-\frac{\pi}{4}\right)$ equals

$\left(\frac{\pi}{4}\right)^{1 / \sqrt{2}}\left(\frac{\sqrt{2}}{2} \ln \frac{\pi}{4}-\frac{2 \sqrt{2}}{\pi}\right)$

$\left(\frac{\pi}{4}\right)^{1 / \sqrt{2}}\left(\frac{\sqrt{2}}{2} \ln \frac{\pi}{4}+\frac{2 \sqrt{2}}{\pi}\right)$

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JEE • MATHEMATICS • DIFFERENTIATION

Differentiation practice · Mixed difficulty

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Showing 631-660 of 932 questions

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631 Medium

631 of 932

Analyzing +4 | -1

Let

y=\ln (1+\cos x)^2

, then the value of

\frac{d^2 y}{d x^2}+\frac{2}{e^{y / 2}}

equals

View answer and explanation

Correct option: A

Option A (0) is correct because after finding the second derivative and simplifying, the expression

\frac{d^2 y}{d x^2}+\frac{2}{e^{y / 2}}

equals 0.

632 Medium

632 of 932

Application +4 | -1

Let

f\left( x \right) = \sin x,\mathrm{g}\left( x \right) = {x^2}

and

h\left( x \right) = {\log _e}x.

F\left( x \right) = \left( {ho\mathrm{g}\,\,of} \right)\left( x \right),

then

F''\left( x \right)

is equal to

View answer and explanation

Correct option: D

- 2\,\mathrm{cose}{\mathrm{c}^2}x

Option D (

- 2\,\mathrm{cose}{\mathrm{c}^2}x

) is correct because it represents the second derivative of the composite function F(x) after applying the chain rule and trigonometric identities.

633 Medium

633 of 932

Application +4 | -1

y = {2^{\log x}}

, then

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

View answer and explanation

Correct option: D

\frac{{{2^{\log x}}.\log 2}}{x}

Option D (

\frac{{{2^{\log x}}.\log 2}}{x}

) is correct because it is the result of applying the chain rule and the derivative of exponential functions.

634 Easy

634 of 932

Application +4 | -1

y = {e^{ax}}\sin \mathrm{b}x,

then

\frac{{{d^2}y}}{{d{x^2}}} - 2a\frac{{dy}}{{dx}} + {a^2}y

is equal to

View answer and explanation

Correct option: C

- {b^2}y

Option C (

- {b^2}y

) is correct because after differentiating the given function twice and substituting back into the given expression, it simplifies to

- {b^2}y

635 Easy

635 of 932

Application +4 | -1

Let

f(x) = x + 3\log (x - 2)

and

g(x) = x + 5\log (x - 1),

then the set of

x

satisfying the inequality

f'(x) < g'(x)

View answer and explanation

Correct option: D

(7/2,\infty )

Option D (

(7/2,\infty )

) is correct because it represents the interval where f'(x) < g'(x), considering the domain restrictions imposed by the logarithmic terms.

636 Hard

636 of 932

Analyzing +4 | -1

Let

S=\left\{(\lambda, \mu) \in R \times R: f(t)=\left(|\lambda| e^{|t|}-\mu\right)\right.

\sin (2|t|), t \in R

, is a differentiable function

\}

. Then

S

is a subest of?

View answer and explanation

Correct option: A

R \times[0, \infty)

Option A is correct because the differentiability condition at t=0 implies that |λ| = μ, which means μ must be non-negative, hence S is a subset of R x [0, ∞).

637 Easy

637 of 932

Application +4 | -1

Let

f

and

g

be differentiable functions satisfying

g^{\prime}(a)=2, g(a)=b

and

fog=I

(identity function). Then,

f^{\prime}(b)

is equal to

View answer and explanation

Correct option: C

\frac{1}{2}

Option C is correct because it correctly applies the chain rule and the properties of inverse functions to find the derivative.

638 Medium

638 of 932

Analyzing +4 | -1

y^2=P(x)

is a polynomial of degree 3 , then

2 \frac{d}{d x}\left(y^3 \frac{d^2 y}{d x^2}\right)=

View answer and explanation

Correct option: C

P(x) P^{\prime \prime \prime}(x)

Option C is correct because by differentiating

y^2 = P(x)

twice and substituting, the expression

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

simplifies to

P(x) P^{\prime \prime \prime}(x)

639 Easy

639 of 932

Application +4 | -1

Let

f(x)=x+\cot x

and

f

is the inverse of

g

, then

g^{\prime}(x)

is equal to

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because using the relationship between a function and its inverse, and differentiating implicitly, we arrive at the given expression.

640 Medium

640 of 932

Application +4 | -1

Let

f\left( x \right) = {x^3}

use Mean value theorem to write

\frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = f'\left( {x + \theta h} \right)

with

0 < \theta < 1

. If

x \ne 0

, then

\lim\limits_{h \to 0} \theta

is equal to

View answer and explanation

Correct option: C

0.5

Option B (

2:1

) is correct because the ratio of the subtangent to the abscissa for the given parabola is 2:1.

641 Easy

641 of 932

Application +4 | -1

It is given that

f^{\prime}(a)

exists, then

\lim\limits _{x \rightarrow a} \frac{x f(a)-a f(x)}{x-a}

is equal to

View answer and explanation

Correct option: A

f(a)-a f^{\prime}(a)

Option A (

f(a)-a f^{\prime}(a)

) is correct because it is the result of evaluating the limit using the definition of the derivative.

642 Easy

642 of 932

Application +4 | -1

g

is the inverse of a function

f

and

f^{\prime}(x)=\frac{1}{1+x^n}

, then

g^{\prime}(x)

is equal to

View answer and explanation

Correct option: B

1+\{g(x)\}^n

Option B (

1+\{g(x)\}^n

) is correct because it directly follows from the inverse function differentiation rule.

643 Easy

643 of 932

Analyzing +4 | -1

(\sin y)^{\sin (\pi x / 2)}-\frac{\sqrt{3}}{2} \operatorname{Sec}^{-1}(2 x)+2^x \tan (\log (x+2))=0

then

d y / d x

x=-1

View answer and explanation

Correct option: C

\frac{3}{\pi \sqrt{\pi^2-3}}

Option C (

\frac{{3}}{{\pi \sqrt{{\pi^2-3}}}}

) is correct because by differentiating the given equation implicitly and substituting x = -1 and solving for dy/dx, we obtain this result.

644 Easy

644 of 932

Application +4 | -1

f\left( x \right) = \left| {{x^2} - 5x + 6} \right|

, then

f'\left( x \right)

equals

View answer and explanation

Correct option: B

5 - 2x

for

2 < x < 3

Option B (

5 - 2x

for

2 < x < 3

) is correct because it represents the derivative of the absolute value function in the interval where the expression inside the absolute value is negative.

645 Easy

645 of 932

Application +4 | -1

x=\phi(t), y=\psi(t)

, then

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

is equal to

View answer and explanation

Correct option: B

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

Option B is correct because it is the result of applying the chain rule and quotient rule to find the second derivative of a parametric function.

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs

646 Easy

646 of 932

Application +4 | -1

Let

f(x)=(a x+b) \cos x+(c x+d) \sin x

and

f^{\prime}(x)=x \cos x

for all

x.

Then,

View answer and explanation

Correct option: A

a=0, b=c=1, d=0

Option A (

a=0, b=c=1, d=0

) is correct because these values satisfy the given condition

f'(x)=x \cos x

when substituted into the derivative of

f(x)

647 Easy

647 of 932

Application +4 | -1

y = x - {x^2}

, then the derivatives of

{y^2}

w.r.t.

{x^2}

View answer and explanation

Correct option: B

2{x^2} - 3x + 1

Option B (

2{x^2} - 3x + 1

) is correct because it represents the derivative of

y^2

with respect to

x^2

, calculated using the chain rule and the given relationship

y = x - x^2

648 Medium

648 of 932

Application +4 | -1

f(x)=x+\tan x

and

f

is inverse of

g

, then

g^{\prime}(x)

is equal to

View answer and explanation

Correct option: C

\frac{1}{2+[g(x)-x]^2}

Option C is correct because it uses the inverse function derivative formula and correctly substitutes the given function.

649 Medium

649 of 932

Analyzing +4 | -1

The set of points where the function

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

is differentiable is

View answer and explanation

Correct option: B

\left( { - \infty ,0} \right) \cup \left( {0,\infty } \right)

Option B (

\left( { - \infty ,0} \right) \cup \left( {0,\infty } \right)

) is correct because the function is not differentiable at x=0 due to the nature of the exponential and square root.

650 Easy

650 of 932

Application +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 of these

Option D (All of these) is correct because all the given relations are satisfied by the functions u and v after differentiation and substitution.

651 Easy

651 of 932

Analyzing +4 | -1

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

is not differentiable at

View answer and explanation

Correct option: C

x=0

Option C (

x=0

) is correct because the function

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

has a sharp corner at

x=0

, making it non-differentiable at that point.

652 Easy

652 of 932

Analyzing +4 | -1

y = \frac{x}{{a + \frac{x}{{b + \frac{x}{{a + \frac{x}{{b + .....\infty }}}}}}}},

then

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

equals

View answer and explanation

Correct option: D

\frac{b}{{ab + 2ay}}

Option D (

\frac{b}{{ab + 2ay}}

) is correct because it is the derivative of y with respect to x after substituting the continued fraction.

653 Easy

653 of 932

Analyzing +4 | -1

f(x)=\cos ^{-1}\left(4 x^3-3 x\right), x \in[-1,1]

then

View answer and explanation

Correct option: D

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

Option D is correct because the derivative of arccos(4x^3-3x) 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 arccos.

654 Easy

654 of 932

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 differentiating the given equation twice and solving for

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

yields this result.

655 Medium

655 of 932

Analyzing +4 | -1

{y^2} = P\left( x \right)

is a polynomial of degree 3, then

2\frac{d}{{dx}}\left[ {{y^3}\frac{{{d^2}y}}{{d{x^2}}}} \right]

equals

View answer and explanation

Correct option: C

P\left( x \right).P'''\left( x \right)

Option C (

P\left( x \right).P'''\left( x \right)

) is correct because after performing the differentiation and simplification, the expression reduces to this form.

656 Easy

656 of 932

Analyzing +4 | -1

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

, then

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

equals

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because after substituting

x = \sin A

and

y = \sin B

, and differentiating implicitly, the expression simplifies to this result.

657 Easy

657 of 932

Application +4 | -1

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

, then

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

equals

View answer and explanation

Correct option: A

\left(\frac{\pi}{4}\right)^{1 / \sqrt{2}}\left(\frac{\sqrt{2}}{2} \ln \frac{4}{\pi}-\frac{2 \sqrt{2}}{\pi}\right)

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

658 Easy

658 of 932

Application +4 | -1

x=\frac{t+1}{t}, y=\frac{t-1}{t}

then

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

t=1

View answer and explanation

Correct option: A

Option A (0) is correct because the second derivative is zero, as the first derivative is a constant.

659 Easy

659 of 932

Application +4 | -1

y=\cos ^{-1}\left(\frac{3 \cos x+4 \sin x}{5}\right)

, then

\frac{d y}{d x}=

View answer and explanation

Correct option: B

Option B (1) is correct because the expression inside the inverse cosine simplifies to a cosine function, allowing for direct differentiation.

660 Easy

660 of 932

Understanding +4 | -1

Use the following table and the fact that

f(x)

is invertible and differentiable everywhere to find

f^{-1}(3)

\begin{array}{ccc}x & f(x) & f^{\prime}(x) \\ 3 & 1 & 7 \\ 6 & 2 & 10 \\ 9 & 3 & 5\end{array}

View answer and explanation

Correct option: B

\frac{1}{5}

Option B (

\frac{{1}}{{5}}

) is correct because it uses the inverse function derivative formula and the given table values.

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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