If $y = {5^x}{x^5},$, then $\frac{{dy}}{{dx}}$ is

${5^x}\left( {{x^5}\log 5 - 5{x^4}} \right)$

If $ y=(\sin x)^{\tan x} $, then $ \frac{d y}{d x} $ is equal to

$ \tan x \cdot(\sin x)^{\tan x-1} \cdot \cos x $

$ (\sin x)^{\tan x} \cdot \sec ^{2} x \cdot \log \sin x $

$ \tan x \cdot(\sin x)^{\tan x-1} $

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

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

$f^{\prime}(x)$ exists for all $x \in[-1,1]$

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

$\frac{x \log x}{(1+\log x)^2}$

If ${x^x}{y^y}{z^z} = c$, then $\frac{{\partial z}}{{\partial x}}$ is equal to

$\left( {\frac{{1 + \log z}}{{1 + \log x}}} \right)$

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JEE • MATHEMATICS • DIFFERENTIATION

Differentiation practice · Mixed difficulty

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361 Easy

361 of 932

Application +4 | -1

y = {5^x}{x^5},

, then

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

View answer and explanation

Correct option: D

{5^x}({x^5}\log 5 + 5{x^4})

Option D (

{5^x}({x^5}\log 5 + 5{x^4}))

is correct because it is the result of applying the product rule and differentiating each term correctly.

362 Medium

362 of 932

Application +4 | -1

The value of

\frac{d}{{dx}}\left( {\left| {x - 1} \right| + \left| {x - 5} \right|} \right)

x = 3

, is

View answer and explanation

Correct option: B

Option B (0) is correct because the derivative of the function at x=3 is 0, as the function is constant in the interval (1,5).

363 Easy

363 of 932

Analyzing +4 | -1

f\left( x \right)

and

g\left( x \right)

are two functions with

g\left( x \right) = x - \frac{1}{x}

and

fo\mathrm{g}\left( x \right) = {x^3} - \frac{1}{{{x^3}}},

then

f'\left( x \right)

View answer and explanation

Correct option: A

3{x^2} + 3

Option A (

3{x^2} + 3

) is correct because by substituting

g(x)

into

f(g(x))

and simplifying, we find

f(x) = x^3 + 3x

, and thus

f'(x) = 3x^2 + 3

364 Easy

364 of 932

Application +4 | -1

y=(\sin x)^{\tan x}

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: A

(\sin x)^{\tan x} \cdot\left(1+\sec ^{2} x \cdot \log \sin x\right)

Option A is correct because it is the result of logarithmic differentiation and simplification.

365 Easy

365 of 932

Application +4 | -1

y = \frac{{\sec x - \tan x}}{{\sec x + \tan x}},

then

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

equals

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because it is the derivative of the given function y with respect to x, obtained after simplification and applying the chain rule.

366 Easy

366 of 932

Analyzing +4 | -1

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

Then,

View answer and explanation

Correct option: B

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

Option B is correct because the derivative of arcsin(2x*sqrt(1-x²)) is piecewise, being 2/sqrt(1-x²) for |x|<1/sqrt(2) and -2/sqrt(1-x²) for 1/sqrt(2)<|x|<1 due to the properties of arcsin.

367 Easy

367 of 932

Application +4 | -1

G\left( x \right) = \sqrt {25 - {x^2}}

, then

\lim\limits_{x \to 1} \frac{{G\left( x \right) - G\left( 1 \right)}}{{x - 1}}

has the value

View answer and explanation

Correct option: A

\frac{1}{{\sqrt {24} }}

Option D (

- 1/5

) is correct because the limit represents the derivative of G(x) evaluated at x=1.

368 Medium

368 of 932

Application +4 | -1

f\left( x \right) = {e^x}\sin x,

then

f''\left( x \right)

is equal to

View answer and explanation

Correct option: B

2{e^x}\cos x

Option B (

2{e^x}\cos x

) is correct because it is the result of taking the second derivative of the given function

f(x) = e^x \sin x

369 Easy

369 of 932

Analyzing +4 | -1

Let

f(x) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{\sin (x - [x])}}{{x - [x]}},}&{x \in ( - 2, - 1)} \\ {\max \,\,\,\left\{ {2x,3\left[ {\left| x \right|} \right]} \right\},}&{\left| x \right| < 1} \\ {1,}&{{\text{other}}\,\,{\text{wise}}} \end{array}} \right.

Where

[t]

denotes greatest integer

\leqslant t.

m

is the number of points where

f

is not continuous and

n

is the number of points where

f

is not differentiable, then the ordered pair

(m,n)

View answer and explanation

Correct option: C

(2, 3)

Option C ((2, 3)) is correct because the function f(x) is discontinuous at two points and not differentiable at three points.

370 Easy

370 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 twice and simplifying,

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

equals a constant value of

\frac{{4ac - {b^2}}}{4}

371 Medium

371 of 932

Application +4 | -1

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

, then

f'\left( x \right)

equals

View answer and explanation

Correct option: C

- \left( {\cos x + \sin x} \right),

for

x \in \left( {0,\pi /4} \right)

Option C (

- \left( {\cos x + \sin x} \right)

, for

x \in \left( {0,\pi /4} \right)

) is correct because it accounts for the absolute value and the interval in which the derivative is being evaluated.

372 Medium

372 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 (

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

) is correct because by differentiating the equation x = g(x) + tan(g(x)) and solving for g'(x), we obtain the given expression.

373 Easy

373 of 932

Application +4 | -1

e^{x-y}=x^y

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: C

\frac{\log x}{(1+\log x)^2}

Option C (

\frac{\log x}{(1+\log x)^2}

) is correct because taking the logarithm and differentiating leads to this result.

374 Medium

374 of 932

Application +4 | -1

The differential coefficient of the function

\left| {x - 1\left| + \right|x - 3} \right|

at the point

x = 2

View answer and explanation

Correct option: B

Option B (0) is correct because at x=2, the function f(x) is constant, hence its derivative is zero.

375 Easy

375 of 932

Analyzing +4 | -1

The set of all points where the function

f(x)=x|x|

is twice differentiable is

View answer and explanation

Correct option: B

(-\infty, 0) \cup(0, \infty)

Option B ((−∞,0)∪(0,∞)) is correct because the second derivative exists everywhere except at x=0, making the function twice differentiable on that interval.

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376 Medium

376 of 932

Application +4 | -1

f(x)=\log _x\{\ln (x)\}

, then

f^{\prime}(x)

x=e

View answer and explanation

Correct option: D

e^{-1}

Option D (

e^{-1}

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

f(x)

x = e

377 Easy

377 of 932

Application +4 | -1

If the greatest value of

\frac{x^2-x+c}{x^2+x+c}

\frac{5}{3}

, then find the value of

c

View answer and explanation

Correct option: A

Option A (4) is correct because solving the inequality derived from the given condition yields c = 4.

378 Medium

378 of 932

Application +4 | -1

The derivative of

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

with respect to

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

View answer and explanation

Correct option: B

Option B (1) is correct because both expressions simplify to

2 \tan^{-1}x

, making their derivatives equal, and the ratio of the derivatives is 1.

379 Easy

379 of 932

Application +4 | -1

f(x)=\sin \left\{\frac{\pi}{3}[x]-x^2\right\}

for

2 < x < 3

and

[x]

denotes the greatest integer less than or equal to

x

, then

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

is equal to

View answer and explanation

Correct option: B

-\sqrt{\pi / 3}

Option B (

-\sqrt{\pi / 3}

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

f(x)

x = \sqrt{\pi / 3}

380 Medium

380 of 932

Analyzing +4 | -1

y(x)=x^{x}, x>0

, then

y^{\prime \prime}(2)-2 y^{\prime}

(2) is equal to:

View answer and explanation

Correct option: A

4\left(\log _{e} 2\right)^{2}-2

Option A is correct. By finding the first and second derivatives of

y(x) = x^x

and substituting

x=2

, we can evaluate

y''(2) - 2y'(2)

381 Easy

381 of 932

Analyzing +4 | -1

f\left( x \right) = \left| {\begin{array}{cccccccccccccccccccc} {{x^3}}&{{x^4}}&{3{x^2}}\\ 1&{ - 6}&4\\ p&{{p^2}}&{{p^3}} \end{array}} \right|

, here

p

is a constant, then

\frac{{{d^4}f\left( x \right)}}{{d{x^4}}}

View answer and explanation

Correct option: D

A constant

The correct answer is 'A constant' because after taking the fourth derivative of the determinant, the resulting expression is a constant value.

382 Medium

382 of 932

Application +4 | -1

x = \log \left( {1 + {t^2}} \right)

and

y = t - {\tan ^{ - 1}}t

.Then

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

is equal to

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because after finding the derivatives and substituting, the expression simplifies to

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

383 Easy

383 of 932

Application +4 | -1

{x^x}{y^y}{z^z} = c

, then

\frac{{\partial z}}{{\partial x}}

is equal to

View answer and explanation

Correct option: B

- \left( {\frac{{1 + \log x}}{{1 + \log z}}} \right)

Option B (

-(\frac{{1 + \log x}}{{1 + \log z}})

) is correct because it is the result of partially differentiating the given equation with respect to x and solving for

\frac{{\partial z}}{{\partial x}}

384 Easy

384 of 932

Application +4 | -1

f(x)=e^x g(x), g(0)=2, g^{\prime}(0)=1

, then

f^{\prime}(0)

View answer and explanation

Correct option: B

Option B (3) is correct because it is the value of the derivative of f(x) at x=0, calculated using the product rule and the given values of g(0) and g'(0).

385 Easy

385 of 932

Analyzing +4 | -1

y = \sin [{\cos ^{ - 1}}\{ \sin ({\cos ^{ - 1}}x)\} ],

then

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

x = \frac{1}{2}

is equal to

View answer and explanation

Correct option: D

Option D (1) is correct because after simplifying the given expression, y becomes equal to x, hence its derivative is 1.

386 Easy

386 of 932

Understanding +4 | -1

Let

f

be the function on

[0,1]

given by

f(x)=

x \sin \frac{\pi}{x}

for

x \neq 0

and

f(0)=0

. Then

f

View answer and explanation

Correct option: D

continuous and bounded

Option D (continuous and bounded) is correct because the function

f(x)

is continuous on the interval

[0,1]

and its absolute value is bounded by

|x|

387 Medium

387 of 932

Application +4 | -1

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

and

f^{\prime}(x)=1-x^2

then

g^{\prime}(x)

equals to :

View answer and explanation

Correct option: C

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

Option C (

2 x(x+\sqrt{1-x^2})

) is correct because it is the derivative of g(x) found by applying the chain rule and simplifying.

388 Medium

388 of 932

Application +4 | -1

y = \sqrt {x + \sqrt {x + \sqrt {x + ...\mathrm{to}\,\,\infty ,} } }

, then

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

is equal to

View answer and explanation

Correct option: D

\frac{1}{{2y - 1}}

Option D (

\frac{1}{{2y - 1}}

) is correct because it is the derivative of the given function after simplifying and differentiating implicitly.

389 Medium

389 of 932

Application +4 | -1

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

then

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

is equal to

View answer and explanation

Correct option: D

- \frac{1}{2}

Option D (-1/2) is correct because simplifying the expression inside the inverse tangent function and then differentiating gives -1/2.

390 Medium

390 of 932

Application +4 | -1

f(x)=\sqrt{\frac{1+\sin ^{-1} x}{1-\tan ^{-1} x}}

; then

f^{\prime}(0)

is equal to :

View answer and explanation

Correct option: D

Option D (1) is correct because the derivative at 0 is found by logarithmic differentiation and substituting x=0.

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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