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

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

If $y = c{e^{x/\left( {x - a} \right)}}$, then $\frac{{dy}}{{dx}}$ equals

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

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JEE • MATHEMATICS • DIFFERENTIATION

Differentiation practice · Medium

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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391 Medium

391 of 402

Analyzing +4 | -1

A curve is given by the equations

x = a\cos \theta + \frac{1}{2}b\cos 2\theta ,y = a\sin \theta + \frac{1}{2}b\sin 2\theta

. Then the points for which

\frac{{{d^2}y}}{{d{x^2}}} = 0

are given by

View answer and explanation

Correct option: C

\cos \theta = - \frac{{{a^2} + 2{b^2}}}{{3ab}}

Option C (

\cos \theta = - \frac{{{{a^2} + 2{{b^2}}}}}{{3ab}}

) is correct because it is the result of setting the second derivative to zero and solving for

\cos \theta

392 Medium

392 of 402

Application +4 | -1

x=2 \cos t-\cos 2 t, y=2 \sin t-\sin 2 t

, then

d y / d x

t=\pi / 6

View answer and explanation

Correct option: D

Option D (1) is correct because, after finding

\frac{dx}{dt}

and

\frac{dy}{dt}

, we compute

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = tan(\frac{3t}{2})

. Evaluating at

t = \frac{\pi}{6}

gives

\frac{dy}{dx} = tan(\frac{\pi}{4}) = 1

393 Medium

393 of 402

Application +4 | -1

\tan \frac{y}{2}=\sqrt{\frac{1-e}{1+e}} \tan \frac{x}{2}

, then

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

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because it is the result of differentiating the given equation and simplifying using trigonometric identities.

394 Medium

394 of 402

Application +4 | -1

u=f\left(x^3\right), v=g\left(x^2\right), f^{\prime}(x)=\cos x

and

g^{\prime}(x)=\sin x

, then

\frac{d u}{d v}

View answer and explanation

Correct option: A

\frac{3}{2} x \cdot \cos x^3 \cdot \operatorname{cosec} x^2

Option A is correct because it accurately applies the chain rule to find du/dv, using the given derivatives of f'(x) and g'(x).

395 Medium

395 of 402

Application +4 | -1

x=\sqrt{2^{\operatorname{cosec}^{-1} t}}

and

y=\sqrt{2^{\sec ^{-1} t}}(|\mathrm{t}| \geq 1)

, then

\frac{d y}{d x}

is equal to.

View answer and explanation

Correct option: B

-\frac{y}{x}

Option B (

-\frac{y}{x}

) is correct because it is the derivative calculated using the relationship between inverse trigonometric functions.

396 Medium

396 of 402

Application +4 | -1

f

and

g

are differentiable functions. If

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

and

f o g=I

, then

f^{\prime}(b)=\ldots \ldots

View answer and explanation

Correct option: A

\frac{1}{2}

Option A (

\frac{{1}}{{2}}

) is correct because it is the value of

f^{\prime}(b)

obtained by differentiating the composite function and using the given values.

397 Medium

397 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 it is the result of differentiating the composite function h(x) and simplifying the expression.

398 Medium

398 of 402

Analyzing +4 | -1

Differential coefficient of

\left(x^{\frac{l+m}{m-n}}\right)^{\frac{1}{n-l}} \cdot\left(x^{\frac{m+n}{n-l}}\right)^{\frac{1}{l-m}} \cdot\left(x^{\frac{n+l}{l-m}}\right)^{\frac{1}{m-n}}

w.r.t.

x

, is

View answer and explanation

Correct option: B

Option B (0) is correct because the expression simplifies to a constant, and the derivative of a constant is zero.

399 Medium

399 of 402

Analyzing +4 | -1

f^{\prime \prime}(x)=-f(x), g(x)=f^{\prime}(x)

F(x)=\left(f\left(\frac{x}{2}\right)\right)^2+\left(g\left(\frac{x}{2}\right)\right)^2

and

F(5)=5

, then

F(10)

is equal to

View answer and explanation

Correct option: A

Option A (5) is correct because

F(x)

is a constant function, so

F(10) = F(5) = 5

400 Medium

400 of 402

Application +4 | -1

y = \sin ({\log _e}x),

then

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

is equal to

View answer and explanation

Correct option: D

- y

Option D (

-y

) is correct because after finding the first and second derivatives and substituting them into the given expression, it simplifies to -y.

401 Medium

401 of 402

Application +4 | -1

y = c{e^{x/\left( {x - a} \right)}}

, then

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

equals

View answer and explanation

Correct option: B

- \frac{{ay}}{{{{\left( {x - a} \right)}^2}}}

Option B (

- \frac{{ay}}{{{{\left( {x - a} \right)}^2}}})

is correct because it is the result of applying the chain rule and quotient rule correctly during differentiation.

402 Medium

402 of 402

Application +4 | -1

f: R \rightarrow R, g: R \rightarrow R

be twice differentiable function satisfying

f^{11}(x)=g^{11}(x)

2 f^1(1)=g^1(1)=4

and

3 f(2)=g(2)=9

then

f(4)-g(4)=

View answer and explanation

Correct option: C

-10

Option C (-10) is correct because it is the value obtained by solving the differential equation

f^{11}(x)=g^{11}(x)

with the given initial conditions and evaluating

f(4)-g(4)

Revision notes and key takeaways

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