If $ \sin (x y)+\frac{x}{y}=x^{2}-y $, then $ \frac{d y}{d x} $ is equal to

$ \frac{\left[2 x y-y^{2} \cos (x y)-1\right]}{x y^{2} \cos (x y)+y^{2}-x} $

If $ x \log _{e}\left(\log _{e} x\right)-x^{2}+y^{2}=4(y>0) $, then $ \frac{d y}{d x} $ at $ \mathrm{x}=\mathrm{e} $ is equal to :

$ \frac{(1+2 e)}{2 \sqrt{4+e^{2}}} $

$ \frac{(1+2 e)}{\sqrt{4+e^{2}}} $

If $\sqrt{1-x^2}+\sqrt{1-y^2}=\sqrt{3}(x-y)$, then $d y / d x$ is equal to

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

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

$\frac{x^2-y^2}{\sqrt{\left(1-x^2\right)\left(1-y^2\right)}}$

Let $f\left( x \right) = {e^x},{\rm{g}}\left( x \right) = {\sin ^{ - 1}}x\,$ and $h\left( x \right) = f\left[ {{\rm{g}}\left( x \right)} \right],$ then $\frac{{h'\left( x \right)}}{{h\left( x \right)}}$ is equal to

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

The derivative of $y = {x^{{\rm{In}}\,x}}$ is

$2{x^{{\rm{In}}\,\,x - 1}}$ In $x$

${x^{{\rm{In}}\,\,x - 1}}$ In $x$

If $x \log _e\left(\log _e x\right)-x^2+y^2=4(y > 0)$, then $d y / d x$ at ${x}={e}$ is equal to:

$\frac{(1+2 {e})}{2 \sqrt{4+{e}^2}}$

$\frac{(1+2 {e})}{\sqrt{4+{e}^2}}$

If ${y}={A} \sin \omega {t}$ then $\frac{d^5 y}{d t^5}=$

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

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

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JEE • MATHEMATICS • DIFFERENTIATION

Differentiation practice · Medium

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301 Medium

301 of 402

Application +4 | -1

f(x)=\left\{\begin{array}{ll}e^x, & x<2 \\ a x+b, & x \geq 2\end{array}\right.

is differentiable for all

x \in R

, then

View answer and explanation

Correct option: A

a=e^2, b=-e^2

Option A is correct because for the function to be differentiable at

x=2

, both continuity and differentiability must hold, leading to

a=e^2

and

b=-e^2

302 Medium

302 of 402

Application +4 | -1

f(x)=\cos x \cos 2 x \cos 2^2 x \cos 2^3 x \ldots \ldots

\cos 2^{n-1} x

and

n>1

, then

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

View answer and explanation

Correct option: A

Option A (1) is correct because by using the identity

\sin(2x) = 2\sin(x)\cos(x)

repeatedly, we can simplify

f(x)

\frac{{\sin(2^n x)}}{{2^n \sin(x)}}

, then differentiate and evaluate at

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

303 Medium

303 of 402

Analyzing +4 | -1

The set of points where the function

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

is differentiable, is

View answer and explanation

Correct option: B

R - \left\{ 1 \right\}

Option B (

R - \\{1\\}

) is correct because the absolute value function

|x - 1|

is not differentiable at

x = 1

, making the product

f(x) = |x - 1|e^x

also non-differentiable at that point.

304 Medium

304 of 402

Application +4 | -1

f(x)=1-x^{2 / 3}+\frac{16}{x}

, then

f^{\prime}(-8)

View answer and explanation

Correct option: C

\frac{1}{12}

Option C (

\frac{{1}}{{12}}

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

f(x)

evaluated at

x = -8

305 Medium

305 of 402

Application +4 | -1

\frac{d}{d x}\left[\frac{e^{a x}}{\sin (b x+c)}\right]

is equal to

View answer and explanation

Correct option: C

\frac{e^{a x}[a \sin (b x+c)-b \cos (b x+c)]}{\sin ^{2}(b x+c)}

Option C is correct because the derivative is found using the quotient rule and simplifying the resulting expression.

306 Medium

306 of 402

Application +4 | -1

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

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because it is the result of implicit differentiation.

307 Medium

307 of 402

Application +4 | -1

x \log _{e}\left(\log _{e} x\right)-x^{2}+y^{2}=4(y>0)

, then

\frac{d y}{d x}

\mathrm{x}=\mathrm{e}

is equal to :

View answer and explanation

Correct option: B

\frac{(2 e-1)}{2 \sqrt{4+e^{2}}}

Option B (

\frac{(2 e-1)}{2 \sqrt{4+e^{2}}}

) is correct because it is the result of differentiating the given equation implicitly with respect to x and substituting x=e and y=

\sqrt{4+e^{2}}

308 Medium

308 of 402

Application +4 | -1

Let

f(x)

be a differentiable function at

x = a

with

f'(a) = 2

and

f(a) = 4

. Then,

\lim\limits_{x \to a} \frac{{xf(a) - af(x)}}{{x - a}}

equals

View answer and explanation

Correct option: B

4 - 2a

Option B (4 - 2a) is correct because it is the result of applying L'Hopital's rule or using the definition of the derivative to evaluate the limit.

309 Medium

309 of 402

Analyzing +4 | -1

\sqrt{1-x^2}+\sqrt{1-y^2}=\sqrt{3}(x-y)

, then

d y / d x

is equal to

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because by substituting x = sin(alpha) and y = sin(beta), and differentiating the resulting expression, we arrive at this result.

310 Medium

310 of 402

Application +4 | -1

Let

f\left( x \right) = {e^x},\mathrm{g}\left( x \right) = {\sin ^{ - 1}}x\,

and

h\left( x \right) = f\left[ {\mathrm{g}\left( x \right)} \right],

then

\frac{{h'\left( x \right)}}{{h\left( x \right)}}

is equal to

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because after finding h(x) and differentiating, the expression

\frac{{h'(x)}}{{h(x)}}

simplifies to

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

311 Medium

311 of 402

Analyzing +4 | -1

The number of points at which the function

f(x)=

|x+1|+|x-1|

is not differentiable is

View answer and explanation

Correct option: A

Option A (2) is correct because the function is not differentiable at the two points where the absolute value expressions change sign, namely x = -1 and x = 1.

312 Medium

312 of 402

Application +4 | -1

The derivative of

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

View answer and explanation

Correct option: C

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

Option C,

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

, is correct because it represents the derivative of the given function after applying a trigonometric substitution and simplifying the expression.

313 Medium

313 of 402

Analyzing +4 | -1

Let

f:[2,4] \rightarrow \mathbb{R}

be a differentiable function such that

\left(x \log _e x\right) f^{\prime}(x)+\left(\log _e x\right) f(x)+f(x) \geq 1, x \in[2,4]

with

f(2)=\frac{1}{2}

and

f(4)=\frac{1}{2}

.

Consider the following two statements :

(A)

f(x) \leq 1

, for all

x \in[2,4]

(B)

f(x) \geq \frac{1}{8}

, for all

x \in[2,4]

Then,

View answer and explanation

Correct option: C

Both the statements (A) and (B) are true

Option D (

\frac{{5}}{{2}}

) is correct because after finding the possible values of α using the given conditions, the sum of these values is equal to 5/2.

314 Medium

314 of 402

Application +4 | -1

Derivative of the function

f\left( x \right) = {\log _5}({\log _7}x),x > 7

View answer and explanation

Correct option: A

\frac{1}{{x(\log 5)(\log 7)({{\log }_7}x)}}

Option A is correct because it is the result of applying the chain rule and the change of base formula for logarithms to find the derivative.

315 Medium

315 of 402

Application +4 | -1

y=|\cos x|+|\sin x|

, then

\frac{d y}{d x}

x=\frac{2 \pi}{3}

View answer and explanation

Correct option: C

\frac{\sqrt{3}-1}{2}

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

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316 Medium

316 of 402

Analyzing +4 | -1

Let

f(x)= \begin{cases}x e^{a x}, & x \leq 0 \\ x+a x^2-x^3, & x>0\end{cases}

where

a

is a positive constant. The interval in which

f^{\prime \prime}(x) \geq 0

, is

View answer and explanation

Correct option: C

\left[\frac{-2}{a}, \frac{a}{3}\right]

Option C ([

\frac{-2}{a}

\frac{a}{3}

]) is correct because it represents the interval where the second derivative of the given piecewise function is greater than or equal to zero.

317 Medium

317 of 402

Application +4 | -1

The derivative of

y = {x^{\mathrm{In}\,x}}

View answer and explanation

Correct option: C

2{x^{\mathrm{In}\,\,x - 1}}

x

Option C (

2{x^{\mathrm{In}\,x - 1}}

x

) is the correct derivative, obtained by logarithmic differentiation.

318 Medium

318 of 402

Application +4 | -1

y=\tan ^{-1}(\operatorname{cosec} x+\cot x)

, then

\frac{d y}{d x}

equals

View answer and explanation

Correct option: A

-\frac{1}{2}

Option A (

-\frac{1}{2}

) is correct because after simplifying the expression inside the arctangent and differentiating, the derivative is found to be

-\frac{1}{2}

319 Medium

319 of 402

Application +4 | -1

x \log _e\left(\log _e x\right)-x^2+y^2=4(y > 0)

, then

d y / d x

{x}={e}

is equal to:

View answer and explanation

Correct option: C

\frac{(2 {e}-1)}{2 \sqrt{4+{e}^2}}

Option C ((2e-1)/(2√(4+e²))) is correct because differentiating the given equation implicitly and substituting x=e allows us to solve for dy/dx, yielding the correct expression.

320 Medium

320 of 402

Application +4 | -1

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

then

\left( {1 + {x^2}} \right)\frac{{{d^2}y}}{{d{x^2}}} + x\frac{{dy}}{{dx}}

View answer and explanation

Correct option: A

{n^2}y

Option A (

{n^2}y

) is correct because after differentiating twice and rearranging, the expression equals

{n^2}y

321 Medium

321 of 402

Application +4 | -1

\frac{d}{d x} \cos ^{-1} \sqrt{\frac{1+x^{2}}{2}}

is equal to

View answer and explanation

Correct option: C

\frac{-x}{\sqrt{1-x^{4}}}

Option C (

\frac{{-x}}{{\sqrt{{1-x^{4}}}}}

) is correct because by substituting

x^2 = \cos 2\theta

, the expression simplifies to

\theta

, and its derivative is

\frac{{-x}}{{\sqrt{{1-x^{4}}}}}

322 Medium

322 of 402

Analyzing +4 | -1

y = \int \limits_0^x f\left( t \right)\sin \left\{ {k\left( {x - t} \right)} \right\}dt

, then

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

equals

View answer and explanation

Correct option: C

kf\left( x \right)

Option D (

k f(x)

) is correct because it is the result of applying Leibniz's rule for differentiating under the integral sign and simplifying the expression.

323 Medium

323 of 402

Application +4 | -1

f(x)=\left|\begin{array}{ll}\sin x & \cos x \\ \tan x & \cot x\end{array}\right|

, then the value of

-256 \sqrt{2} f^{\prime}\left(\frac{\pi}{4}\right)

must be:

View answer and explanation

Correct option: A

512

Option A (512) is correct because it is the value obtained after calculating the derivative of the determinant and substituting x = π/4.

324 Medium

324 of 402

Application +4 | -1

y = {\cos ^{ - 1}}\left( {\frac{{2\cos x - 3\sin x}}{{\sqrt {13} }}} \right)

, then

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

View answer and explanation

Correct option: B

Constant =1

Option B (Constant =1) is correct because after simplification, the derivative of y with respect to x is equal to 1.

325 Medium

325 of 402

Application +4 | -1

y = \log (\sin \,({x^2})),0 < x < \frac{\pi }{2},

then

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

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

View answer and explanation

Correct option: D

\sqrt \pi

Option D (

\sqrt{\pi}

) is correct because it is the value obtained after differentiating the given function and substituting the given value of x.

326 Medium

326 of 402

Application +4 | -1

\sqrt{x+y}+\sqrt{y-x}=c

then

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

equals

View answer and explanation

Correct option: A

\frac{2}{c^2}

Option A is correct because the second derivative is found by implicit differentiation of the given equation, resulting in a constant value.

327 Medium

327 of 402

Application +4 | -1

If the function

f\left( x \right)

is defined by

f\left( x \right) = a + bx

and

{f^r} = fff \ldots

(repeated

r

times), then

{f^r}\left( x \right)

is equal to

View answer and explanation

Correct option: D

a\left( {\frac{{{b^r} - 1}}{{b - 1}}} \right) + {b^r}x

Option D,

a\left( {\frac{{{{b^r} - 1}}{{{{b - 1}}}}} \right) + {{b^r}x}

, is correct because it represents the general form of the function f(x) repeated r times, derived by iteratively applying the function to itself.

328 Medium

328 of 402

Application +4 | -1

Let

y(x)=e^{2 \sin ^{-1} x}, x \in[-1,1]

and

\left(1-x^2\right) y^{\prime \prime}(x)=x y^{\prime}(x)+\lambda y(x)

, then

\lambda

equals

View answer and explanation

Correct option: A

Option A (4) is correct. By differentiating the given function twice and substituting into the given differential equation, the value of \

\lambda

\$ is found to be 4.

329 Medium

329 of 402

Application +4 | -1

If a variable tangent to the curve

x^2 y=c^3

makes intercepts

a, b

x

and

y

axis respectively, then the value of

a^2 b

View answer and explanation

Correct option: C

\frac{27}{4} c^3

Option C (

\frac{27}{4} c^3

) is correct because it is the value of

a^2b

calculated from the given curve equation and tangent properties.

330 Medium

330 of 402

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 is correct because the fifth derivative of y = A sin(ωt) is Aω^5 sin(ωt + π/2), which can be found by repeatedly differentiating the function.

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