If $ x=a(t-\sin t) $ and $ y=a(1-\cos t) $, then $ \frac{d y}{d x} $ is equal to

$ \tan \left(\frac{t}{2}\right) $

$ -\tan \left(\frac{t}{2}\right) $

If for $ x \in\left(0, \frac{1}{4}\right) $, the derivative of $ \tan ^{-1}\left(\frac{6 x \sqrt{x}}{1-9 x^{3}}\right) $ is $ \sqrt{\mathrm{x}} \cdot \mathrm{g}(\mathrm{x}) $, then $ g(x) $ equals :

$ \frac{3 x \sqrt{x}}{1-9 x^{3}} $

$ \frac{3 \mathrm{x}}{1-9 \mathrm{x}^{3}} $

If $ y=\frac{e^{x} \log x}{x^{2}} $, then $ \frac{d y}{d x} $ is equal to

$ \frac{e^{x}[1-(x-2) \log x]}{x^{4}} $

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 · Mixed difficulty

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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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Showing 901-930 of 932 questions

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901 Easy

901 of 932

Application +4 | -1

x=a(t-\sin t)

and

y=a(1-\cos t)

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: C

\cot \left(\frac{t}{2}\right)

Option C (

\cot \left(\frac{t}{2}\right)

) is correct because it is the simplified form of

\frac{dy}{dx}

after applying trigonometric identities.

902 Easy

902 of 932

Application +4 | -1

The length

l

of a rectangle is decreasing at the rate of

2 \mathrm{~cm} / \mathrm{sec}

while the widths

w

is increasing at the rate of

2 \mathrm{~cm} / \mathrm{sec}

. When

l=12

and

w=5

, the rate of change of area is (in

\mathrm{cm}^2 / \mathrm{sec}

)

View answer and explanation

Correct option: A

Option A (14) is correct because it is the rate of change of the area calculated using the product rule of differentiation.

903 Medium

903 of 932

Application +4 | -1

\frac{d}{d x}\left(\tan ^{-1} \frac{\cos x}{1+\sin x}\right)

is equal to

View answer and explanation

Correct option: A

-\frac{1}{2}

The correct answer is A (

-\frac{1}{2}

) because after simplifying the trigonometric expression inside the inverse tangent, the derivative is a constant.

904 Easy

904 of 932

Analyzing +4 | -1

Let

\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}

be a function such that

|f(x)| \leq x^{2}

, for all

x \in R

. Then, at

x=0

\mathrm{f}

is:

View answer and explanation

Correct option: B

continuous as well as differentiable.

Option B is correct because the given condition |f(x)| ≤ x² implies that f(0) = 0 and f'(0) = 0, making the function both continuous and differentiable at x=0.

905 Easy

905 of 932

Application +4 | -1

{x^p}{y^q} = {\left( {x + y} \right)^{p + q}}

, then

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

is equal to

View answer and explanation

Correct option: A

\frac{y}{x}

Option A,

\frac{y}{x}

, is correct because it is the simplified result of differentiating the given equation

{x^p}{y^q} = {\left( {x + y} \right)^{p + q}}

implicitly with respect to x.

906 Easy

906 of 932

Application +4 | -1

If for

x \in\left(0, \frac{1}{4}\right)

, the derivative of

\tan ^{-1}\left(\frac{6 x \sqrt{x}}{1-9 x^{3}}\right)

\sqrt{\mathrm{x}} \cdot \mathrm{g}(\mathrm{x})

, then

g(x)

equals :

View answer and explanation

Correct option: B

\frac{9}{1+9 x^{3}}

Option B (

\frac{9}{1+9 x^{3}}

) is correct because it is the expression for g(x) after simplifying and differentiating the given function.

907 Medium

907 of 932

Application +4 | -1

5f\left( x \right) + 3f\left( {\frac{1}{x}} \right) = x + 2

and

y = x\mathrm{ }f\left( x \right),

then

{\left( {\frac{{dy}}{{dx}}} \right)_{x = 1}}

is equal to

View answer and explanation

Correct option: B

\frac{7}{8}

Option B (

\frac{7}{8}

) is correct because solving the system of equations and differentiating yields the correct result at x=1.

908 Medium

908 of 932

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

909 Easy

909 of 932

Application +4 | -1

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

, where

f(x)

is a continuous double differentiable function and

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

.

If

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)

View answer and explanation

Correct option: B

Option B (5) is correct because F(x) is a constant function, and since F(5) = 5, then F(10) = 5.

910 Easy

910 of 932

Application +4 | -1

The derivative of

\left(x^2+1\right) \cos x

is given by

View answer and explanation

Correct option: A

-x^2 \sin x-\sin x+2 x \cos x

Option A (

-x^2 \sin x - \sin x + 2x \cos x

) is correct because it is the result of applying the product rule to differentiate the given function.

911 Medium

911 of 932

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

912 Medium

912 of 932

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.

913 Easy

913 of 932

Understanding +4 | -1

f:R \to R

is an even function having derivatives of all orders, then an odd function among the following is

View answer and explanation

Correct option: B

f'''

Option B (

f'''

) is correct because the third derivative of an even function is an odd function.

914 Easy

914 of 932

Analyzing +4 | -1

For

x \in R, f(x)=|\log 2-\sin x|

and

g(x)=f(f(x))

, the

View answer and explanation

Correct option: B

g^{\prime}(0)=\cos (\log 2)

Option B (

g^{\prime}(0)=\cos (\log 2)

) is correct because after analyzing the function

g(x)

near

x=0

and applying the chain rule, the derivative at

x=0

simplifies to

\cos(\log 2)

915 Easy

915 of 932

Application +4 | -1

y

is a function of

x

and

\log (x+y)=2 x y

, then the value of

y^{\prime}(0)

View answer and explanation

Correct option: A

Option A (1) is correct because by differentiating the given equation and substituting x=0 and y=1, we find y'(0) = 1.

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916 Medium

916 of 932

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).

917 Easy

917 of 932

Application +4 | -1

y=\frac{e^{x} \log x}{x^{2}}

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: D

\frac{e^{x}[1+(x-2) \log x]}{x^{3}}

Option D is correct because the derivative is found using the product and quotient rules, simplifying to the given expression.

918 Medium

918 of 932

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.

919 Easy

919 of 932

Understanding +4 | -1

y=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\cdots \infty

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: A

y

Option A (

y

) is correct because the derivative of the given series, which represents

e^x

, is equal to the original series.

920 Medium

920 of 932

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.

921 Easy

921 of 932

Application +4 | -1

A point on the curve

y=x^3-3 x+5

at which the tangent line is parallel to

y=-2 x

View answer and explanation

Correct option: C

\left(\frac{1}{\sqrt{3}}, 5-\frac{8 \sqrt{3}}{9}\right)

Option C

\left(\frac{{1}}{{\sqrt{{3}}}}, 5-\frac{{8 \sqrt{{3}}}}{{9}}\right)

is correct because at this point, the derivative of the curve equals the slope of the given line.

922 Medium

922 of 932

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.

923 Easy

923 of 932

Application +4 | -1

The derivative of

{e^x}^3

with respect to

\log x

View answer and explanation

Correct option: C

3{x^3}{e^x}^3

Option C (

3{x^3}{e^x}^3

) is correct because it is the result of correctly applying the chain rule and dividing the derivatives.

924 Medium

924 of 932

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.

925 Easy

925 of 932

Application +4 | -1

y=x \log x-x

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: C

\log x

Option C (

\log x

) is correct because the derivative of

x \log x - x

with respect to

x

\log x

926 Medium

926 of 932

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

927 Hard

927 of 932

Analyzing +4 | -1

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

here

p

is a constant, then

\frac{{{d^3}f\left( x \right)}}{{d{x^3}}}

View answer and explanation

Correct option: D

A constant

Option D (A constant) is correct because after taking the third derivative of the determinant function, the result is a constant value.

928 Medium

928 of 932

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.

929 Medium

929 of 932

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.

930 Medium

930 of 932

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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