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 $f\left( x \right) = \sqrt {{x^2} - 2x + 1} $, then

$f'\left( x \right) = 1$ for all $x$

$f'\left( x \right) = - 1$ for all $x \le 1$

$f'\left( x \right) = 1$ for all $x \ge 1$

If $y^2+x^2=R^2$ and $k=1 / R$, then $k$ is equal to

$\frac{\left|y^{\prime \prime}\right|}{\sqrt{1+y^{\prime 2}}}$

$\frac{2\left|y^{\prime \prime}\right|}{\sqrt{1+y^{\prime 2}}}$

$\frac{\left|y^{\prime \prime}\right|}{2 \sqrt{\left(1+y^{\prime 2}\right)^3}}$.

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JEE • MATHEMATICS • DIFFERENTIATION

Differentiation practice · Medium

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Showing 271-300 of 402 questions

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271 Medium

271 of 402

Application +4 | -1

f^{\prime}(x)=\frac{\cos (\log x)}{x}, f^{\prime}(1)-0

and

y=f\left(\frac{2 x+3}{3-2 x}\right)

, then

\frac{d y}{d x}=\ldots \ldots

View answer and explanation

Correct option: C

\frac{12}{(3-2 x)^2} \sin \left(\log \left(\frac{2 x+3}{3-2 x}\right)\right)

Option C is correct because it involves finding the derivative using the chain rule and the given derivative of f(x).

272 Medium

272 of 402

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.

273 Medium

273 of 402

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.

274 Medium

274 of 402

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.

275 Medium

275 of 402

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)

276 Medium

276 of 402

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.

277 Medium

277 of 402

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.

278 Medium

278 of 402

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.

279 Medium

279 of 402

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.

280 Medium

280 of 402

Analyzing +4 | -1

Let

S=\left\{t \in R: f(x)=|x-\pi|\left(e^{|x|}-1\right) \sin |x|\right.

is not differentiable at

t

. Then, the set

S

is equal to

View answer and explanation

Correct option: D

\phi

Option D (

\phi

) is correct because the function

f(x)

is differentiable at both

x=0

and

x=\pi

, meaning the set

S

of non-differentiable points is empty.

281 Medium

281 of 402

Analyzing +4 | -1

The set of points where

f(x)=\frac{x}{1+|x|}

is differentiable is

View answer and explanation

Correct option: C

(-\infty, \infty)

Option C is correct because the function is differentiable for all real numbers, as the derivative exists everywhere.

282 Medium

282 of 402

Application +4 | -1

The value of

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

x = \frac{\pi }{2}

, where

y

is given by

y = {x^{\sin x}} + \sqrt x

, is

View answer and explanation

Correct option: A

1 + \frac{1}{{\sqrt {2\pi } }}

Option A (

1 + \frac{1}{{\sqrt {2\pi } }}

) is correct because it is the value of dy/dx at x=pi/2, calculated using logarithmic differentiation.

283 Medium

283 of 402

Application +4 | -1

x^{m } y^n=(x+y)^{n+1}

, then

\frac{d y}{d x}=\ldots \ldots

View answer and explanation

Correct option: A

\frac{y}{x}

Option A (

\frac{y}{x}

) is correct because differentiating the given equation using logarithms and simplifying leads to dy/dx = y/x.

284 Medium

284 of 402

Application +4 | -1

\sec \left( {\frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}}} \right) = {e^a},

then

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

is equal to

View answer and explanation

Correct option: B

\frac{y}{x}

Option B (y/x) is correct because implicit differentiation and simplification leads to dy/dx = y/x.

285 Medium

285 of 402

Analyzing +4 | -1

f\left( x \right) = \sin x

and

g\left( x \right) = {\mathop\mathrm{s}\nolimits} \,gn\sin x

, then

g'\left( 1 \right)

equals

View answer and explanation

Correct option: C

\cos 1

Option C (

\cos 1

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

g(x)

x=1

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286 Medium

286 of 402

Analyzing +4 | -1

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

, then

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the derivative is 1 for x > 1 and -1 for x < 1, and it's not defined at x=1.

287 Medium

287 of 402

Analyzing +4 | -1

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

, then

f\left( x \right)

is differentiable on

View answer and explanation

Correct option: B

R - \left\{ { - 1,1} \right\}

Option B (

R - \\{-1, 1\\}

) is correct because the derivative of

f(x)

exists for all real numbers except

x = -1

and

x = 1

, where the function is not differentiable.

288 Medium

288 of 402

Application +4 | -1

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

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: B

-\frac{1}{2}

Option B (

-\frac{1}{2}

) is correct because after simplifying the expression using trigonometric identities, the derivative is found to be -1/2.

289 Medium

289 of 402

Application +4 | -1

x = a\left( {\mathrm{cos\theta } + \mathrm{\theta sin\theta }} \right)

and

y = a(\mathrm{sin\theta } - \theta \cos \theta ),

then

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

is equal to

View answer and explanation

Correct option: B

\mathrm{tan}\,\theta

Option B (

\mathrm{tan}\,\theta

) is correct because it is the result of dividing

\frac{dy}{d\theta}

\frac{dx}{d\theta}

290 Medium

290 of 402

Application +4 | -1

The derivative of

\sin \left( {{x^3}} \right)

w.r.t.

\cos \left( {{x^3}} \right)

View answer and explanation

Correct option: C

- \cot \left( {{x^3}} \right)

Option C is correct because it uses the chain rule to find the derivative of both functions and then divides them.

291 Medium

291 of 402

Application +4 | -1

f(1)=3, f^{\prime}(1)=-\frac{1}{3}

, then the derivative of

\left|x^{11}+f(x)\right|^{-2}

x=1

, is

View answer and explanation

Correct option: D

f^{\prime}(1)

Option D (

f^{\prime}(1)

) is correct because after applying the chain rule and substituting the given values, the derivative simplifies to

f^{\prime}(1)

292 Medium

292 of 402

Application +4 | -1

Let

f

and

g

be differentiable functions on

R

(the set of all real numbers) such that

g(1)=2=g^{\prime}(1)

and

f^{\prime}(0)=4

. If

h(x)=f(2 x g(x)+\cos \pi x-3)

then

h^{\prime}(1)

is equal to

View answer and explanation

Correct option: C

Option C (32) is correct because it is the value of

h'(1)

found by applying the chain rule and substituting the given values.

293 Medium

293 of 402

Application +4 | -1

f(2)=4

and

f^{\prime}(2)=1

, then

\lim\limits_{x \to 2} \frac{{x\,f(2) - 2f(x)}}{{x - 2}} =

View answer and explanation

Correct option: A

Option A (2) is correct because it is the result of evaluating the limit using the definition of the derivative and the given values.

294 Medium

294 of 402

Application +4 | -1

y = {a^x}.{b^{2x - 1}},

then

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

View answer and explanation

Correct option: D

y{(\log a{b^2})^2}

Option D (

y{(\log a{b^2})^2}

) is correct because it is the result of taking the second derivative of the given function using logarithmic differentiation.

295 Medium

295 of 402

Application +4 | -1

Let

f(x)=1+\sin (3 x) \cdot g(x)

where

g(x)

is differentiable, then

f^{\prime}(0)

View answer and explanation

Correct option: B

3 g(0)

Option B (

3 g(0)

) is correct because the derivative

f'(x)

is calculated using the product rule and then evaluated at

x=0

296 Medium

296 of 402

Analyzing +4 | -1

The value of

\log _e 2 \frac{d}{d x}\left(\log _{\cos x} \operatorname{cosec} x\right)

x=\frac{\pi}{4}

View answer and explanation

Correct option: D

4

Option D (4) is correct because it involves finding the derivative of the given logarithmic expression and evaluating it at

x=\frac{\pi}{4}

297 Medium

297 of 402

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: C

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

The correct answer is C (

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

) because it is the derivative of y with respect to x using the chain rule.

298 Medium

298 of 402

Application +4 | -1

y^2+x^2=R^2

and

k=1 / R

, then

k

is equal to

View answer and explanation

Correct option: B

\frac{\left|y^{\prime \prime}\right|}{\sqrt{\left(1+y^{\prime 2}\right)^3}}

Option B (

\frac{{\left|y^{\prime \prime}\right|}}{{\sqrt{{\left(1+y^{\prime 2}\right)^3}}}}

) is correct because it represents the curvature (and hence the reciprocal of the radius) of the given circle.

299 Medium

299 of 402

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'(\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 given function at the specified point, calculated using the chain rule.

300 Medium

300 of 402

Analyzing +4 | -1

Let a function

y = y\left( x \right)

be defined parametrically by

x = 2t - \left| t \right|,y = {t^2} + t\left| t \right|

. Then,

y'\left( x \right),x > 0

View answer and explanation

Correct option: D

Does not exist

Option D (Does not exist) is correct because the derivative does not exist at x=0 due to different limits from the left and right.

Revision notes and key takeaways

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