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

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

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

If ${x^y} = {e^{x - y}}$, then $\frac{{dy}}{{dx}}$ is equal to

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

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

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JEE • MATHEMATICS • DIFFERENTIATION

Differentiation practice · Mixed difficulty

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Showing 481-510 of 932 questions

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481 Medium

481 of 932

Application +4 | -1

y = {\sin ^{ - 1}}\left( {\frac{{5x + 12\sqrt {1 - {x^2}} }}{{13}}} \right)

, then

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

is equal to

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because after simplification using trigonometric identities, the derivative can be easily found.

482 Medium

482 of 932

Application +4 | -1

xy = {\tan ^{ - 1}}\left( {xy} \right) + {\cot ^{ - 1}}\left( {xy} \right)

, then

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

is equal to

View answer and explanation

Correct option: B

- \frac{y}{x}

Option B (

- \frac{y}{x}

) is correct because after using the identity

\tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}

, the derivative simplifies to

- \frac{y}{x}

483 Easy

483 of 932

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.

484 Easy

484 of 932

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.

485 Easy

485 of 932

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)

486 Easy

486 of 932

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.

487 Easy

487 of 932

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.

488 Easy

488 of 932

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.

489 Easy

489 of 932

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.

490 Medium

490 of 932

Analyzing +4 | -1

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

where

f\left( x \right)

is a polynomial of degree 3, then

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because when we solve for the coefficients of the cubic polynomial, all the given relations hold true.

491 Medium

491 of 932

Application +4 | -1

x=\frac{3 t}{1+t^2}, y=\frac{3 t^2}{1+t^2}

, then

d y / d x

t=2

View answer and explanation

Correct option: C

-\frac{4}{3}

Option C (

-\frac{{4}}{{3}}

) is correct because it is the value of

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

calculated using parametric differentiation at

t=2

492 Easy

492 of 932

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.

493 Easy

493 of 932

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.

494 Easy

494 of 932

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}}

495 Easy

495 of 932

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.

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496 Easy

496 of 932

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.

497 Easy

497 of 932

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.

498 Medium

498 of 932

Application +4 | -1

y = {\tan ^{ - 1}}(\sec x - \tan x),

then

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

View answer and explanation

Correct option: D

- \frac{1}{2}

Option D (

- \frac{1}{2}

) is correct because after simplifying the expression inside the inverse tangent and differentiating, the result is

- \frac{1}{2}

499 Easy

499 of 932

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.

500 Medium

500 of 932

Application +4 | -1

The derivative of

f\left( x \right) = 3\left| {2 + x} \right|

at the point

{x_0} = - 3

, is

View answer and explanation

Correct option: B

- 3

Option B (

- 3

) is correct because the derivative of

f(x) = 3|2 + x|

x_0 = -3

is -3, as the function is defined as

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

for

x < -2

501 Medium

501 of 932

Application +4 | -1

Let

f

and

g

be differentiable functions and

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

and

g(x)=-f(x)

and let

T(x)=(f(x))^2-(g(x))^2 \quad T^{\prime}(x)=

View answer and explanation

Correct option: D

6 f(x) g(x)

Option D (

6 f(x) g(x)

) is correct because it is the result of differentiating

T(x)

using the product rule and substituting the given derivatives of

f(x)

and

g(x)

502 Easy

502 of 932

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.

503 Easy

503 of 932

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.

504 Easy

504 of 932

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.

505 Medium

505 of 932

Application +4 | -1

{x^y} = {e^{x - y}}

, then

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

is equal to

View answer and explanation

Correct option: C

\log x.{\left( {1 + \log x} \right)^{ - 2}}

Option C,

\log x.{\left( {1 + \log x} \right)^{ - 2}}

, is correct because it represents the derivative dy/dx obtained by implicit differentiation and simplification of the given equation

{x^y} = {e^{x - y}}

506 Medium

506 of 932

Application +4 | -1

y=(\log x)^x+x^{\log x}

, then

\frac{d y}{d x}=

View answer and explanation

Correct option: A

(\log x)^x\left[\log (\log x)+\frac{1}{\log x}\right]+x^{\log x} \cdot\left(\frac{2 \log x}{x}\right)

Option A is correct because it accurately applies the chain rule and logarithmic differentiation to find the derivative of the given function.

507 Easy

507 of 932

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.

508 Medium

508 of 932

Application +4 | -1

Let

y = {x^{{x^{x \ldots }}}}

, then

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

is equal to

View answer and explanation

Correct option: B

\frac{{{y^2}}}{{x(1 - y\log x)}}

Option B (

\frac{{{y^2}}}{{x(1 - y\log x)}}

) is correct because it results from applying logarithmic differentiation to the given function and solving for

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

509 Easy

509 of 932

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.

510 Medium

510 of 932

Understanding +4 | -1

f\left( x \right)

is a polynomial of degree

n\left( { > 2} \right)

and

f\left( x \right) = f\left( {\alpha - x} \right)

, (where

\alpha

is fixed real number), then the degree of

f'\left( x \right)

View answer and explanation

Correct option: B

n - 1

Option B (

n - 1

) is correct because the derivative of a polynomial of degree

n

is a polynomial of degree

n-1

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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