If $x=f(t)$ and $y=g(t)$, then $\frac{d^2 y}{d x^2}$ is equal to

$\frac{g^{\prime \prime}(t)}{f^{\prime \prime}(t)}$

If $u = - f''\,\,\left( {\rm{\theta }} \right)\,\,\sin \,\,{\rm{\theta }}\,\, + \,\,f'\,\,\left( {\rm{\theta }} \right)\,\,\cos \,\,{\rm{\theta }}$ and $v = f''\,\,\left( {\rm{\theta }} \right)\,\,\cos \,\,{\rm{\theta }}\,\, + \,\,f'\,\,\left( {\rm{\theta }} \right)\,\,\sin \,\,{\rm{\theta }}$, then $f\,\,{\left[ {{{\left( {\frac{{du}}{{d{\rm{\theta }}}}} \right)}^2} + \,\,{{\left( {\frac{{dv}}{{d{\rm{\theta }}}}} \right)}^2}} \right]^{\frac{1}{2}}}d{\rm{\theta }}$ is equal to

$f\,\,\left( {\rm{\theta }} \right)\,\, + \,\,f''\,\,\left( {\rm{\theta }} \right)\,\, + \,\,c$

$f\,\,\left( {\rm{\theta }} \right)\,\, - \,\,f'\,\,\left( {\rm{\theta }} \right)\,\, + \,\,c$

$f'\,\,\left( {\rm{\theta }} \right)\,\, + \,\,f''\,\,\left( {\rm{\theta }} \right)\,\, + \,\,c$

$f'\,\,\left( {\rm{\theta }} \right)\,\, - \,\,f''\,\,\left( {\rm{\theta }} \right)\,\, + \,\,c$

If $ y \sec x+\tan x+x^{2} y=0 $, then $ \frac{d y}{d x} $ is equal to

$ -\frac{2 x y+\sec ^{2} x+\sec x \tan x}{x^{2}+\sec x} $

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JEE • MATHEMATICS • DIFFERENTIATION

Differentiation practice · Mixed difficulty

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Practice format and chapter quiz

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 61-90 of 932 questions

61 Medium

61 of 932

Application +4 | -1

f^{\prime}(1)=2

and

g^{\prime}(\sqrt{2})=4

, then the derivative of

f(\tan x)

w.r.t.

g(\sec x)

x=\frac{\pi}{4}

View answer and explanation

Correct option: C

\sqrt{2} / 2

Option C (

\sqrt{2} / 2

) is correct because it is the result of differentiating

f(\tan x)

with respect to

g(\sec x)

and evaluating at

x = \frac{\pi}{4}

62 Easy

62 of 932

Application +4 | -1

f(x)=\tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right), x \in R-\left\{\frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right\}

, then

View answer and explanation

Correct option: A

f^{\prime}(x)=\frac{3}{1+x^2}

for all

x \in R-\left\{\frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right\}

Option A (

f^{\prime}(x)=\frac{{3}}{{1+x^2}}

for all

x \in R-\{\frac{{-1}}{{\sqrt{3}}}, \frac{{1}}{{\sqrt{3}}}\})

is correct because the derivative of the given function is

\frac{{3}}{{1+x^2}}

everywhere except at the points where the function is not defined.

63 Easy

63 of 932

Application +4 | -1

x=f(t)

and

y=g(t)

, then

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

is equal to

View answer and explanation

Correct option: B

\frac{g^{\prime \prime}(t) f^{\prime}(t)-g^{\prime}(t) f^{\prime \prime}(t)}{\left(f^{\prime}(t)\right)^3}

Option B (

\frac{g^{\prime \prime}(t) f^{\prime}(t)-g^{\prime}(t) f^{\prime \prime}(t)}{\left(f^{\prime}(t)\right)^3}

) is correct because it is derived by applying the chain rule and quotient rule to find the second derivative of y with respect to x.

64 Medium

64 of 932

Analyzing +4 | -1

f(x)=x^{3}-x^{2} f^{\prime}(1)+x f^{\prime \prime}(2)-f^{\prime \prime}(3), x \in R

, then

View answer and explanation

Correct option: A

2 f(0)-f(1)+f(3)=f(2)

Option A (

2 f(0)-f(1)+f(3)=f(2)

) is correct because after finding the derivatives and solving for the constants, the equation holds true.

65 Easy

65 of 932

Analyzing +4 | -1

x^2+y^2=1+\frac{1}{t}

and

x^4+y^4=t^2+\frac{1}{t^2}

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: B

-\frac{y}{x}

Option B (-y/x) is correct because after squaring the first equation and substituting the second, we can find a relationship between x and y, and then differentiate implicitly.

66 Easy

66 of 932

Application +4 | -1

f\left( x \right) - {\log _e}\left( {{{\log }_e}x} \right)

, then

f'\left( x \right)

x = e

, is

View answer and explanation

Correct option: C

\frac{1}{e}

Option C (

\frac{1}{e}

) is correct because after applying the chain rule and simplifying, the derivative at x=e is

\frac{1}{e}

67 Easy

67 of 932

Application +4 | -1

y = {e^{\left( {\frac{1}{2}} \right)\log (1 + {{\tan }^2}x)}}

, then

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

is equal to

View answer and explanation

Correct option: C

\sec x\tan x

Option C (

\sec x\tan x

) is correct because after simplifying the given equation, y = sec(x), and its derivative is sec(x)tan(x).

68 Medium

68 of 932

Application +4 | -1

f\left( {x + y} \right) = f\left( x \right)f\left( y \right)

for all

x,y \in R,f\left( 5 \right) = 2,f'\left( 0 \right) = 3

. Then

f'\left( 5 \right)

equals

View answer and explanation

Correct option: A

Option A (6) is correct because by using the given functional equation and derivative information, we can find f'(5).

69 Easy

69 of 932

Analyzing +4 | -1

Let

f(x)

be the function given by

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

. Then,

View answer and explanation

Correct option: B

f^{\prime}(x)= \begin{cases}\frac{2}{1+x^2}, & x>0 \\ \frac{-2}{1+x^2}, & x<0\end{cases}

Option B is correct because the derivative of arccos((1-x²)/(1+x²)) is piecewise, being 2/(1+x²) for x>0 and -2/(1+x²) for x<0 due to the properties of arctangent.

70 Easy

70 of 932

Analyzing +4 | -1

x = \frac{{2at}}{{1 + {t^3}}}

and

y = \frac{{2a{t^2}}}{{{{\left( {1 + {t^3}} \right)}^2}}}

, then

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

View answer and explanation

Correct option: C

\frac{x}{a}

The correct answer is C (

\frac{x}{a}

) because the relationship between x and y allows for a direct substitution and differentiation.

71 Easy

71 of 932

Analyzing +4 | -1

Let

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

. Then,

View answer and explanation

Correct option: D

All the above

Option D (All the above) is correct because the polynomial

f(x) = x^3 - 5x^2 + 2x + 6

satisfies all the given conditions:

f(0)+f(2)=f(1)

f(0)+f(3)=0

, and

f(1)+f(3)=f(2)

72 Easy

72 of 932

Application +4 | -1

y=y(x)

is a solution of the differential equation

(x+1)^4 d y+(2 x+2) \sin x d x=\left(x^2+1+2 x\right) \cos x d x

such that

y(0)=0

, then

y\left(\frac{\pi}{2}\right)

is equal to-

View answer and explanation

Correct option: D

\frac{4}{(\pi+2)^2}

Option D (

\frac{4}{(\pi+2)^2}

) is correct because solving the differential equation and applying the initial condition yields

y(x) = \frac{\sin x}{(x+1)^2}

, and thus

y(\frac{\pi}{2}) = \frac{4}{(\pi+2)^2}

73 Medium

73 of 932

Application +4 | -1

y = {\log _u}\left| {\cos 4x} \right| + \left| {\sin x} \right|

, where

u = \sec 2x

, then

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

x = - \frac{\pi }{6}

is equal to

View answer and explanation

Correct option: A

\frac{{ - 6\sqrt 3 }}{{{{\log }_e}2}} - \frac{{\sqrt 3 }}{2}

Option A (

\frac{{ - 6\sqrt 3 }}{{{{\log }_e}2}} - \frac{{\sqrt 3 }}{2}

) is correct because it is the result of differentiating the given function and substituting

x = -\frac{\pi}{6}

74 Easy

74 of 932

Analyzing +4 | -1

Let

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

and

g\left( x \right) = - {2^x} + 2x\log 2

. Then the set of points satisfying

f'\left( x \right) > g'\left( x \right)

View answer and explanation

Correct option: C

\left( {0,\infty } \right)

Option C (

\left( {0,\infty } \right)

) is correct because the inequality

f'\left( x \right) > g'\left( x \right)

simplifies to

x > 0

75 Medium

75 of 932

Application +4 | -1

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

and

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

then

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

is equal to

View answer and explanation

Correct option: A

\frac{{12}}{{9 - 4{x^2}}}\cos \left\{ {\log \left( {\frac{{2x + 3}}{{3 - 2x}}} \right)} \right\}

Option A (

\frac{{12}}{{9 - 4{x^2}}}\cos \left\{ {\log \left( {\frac{{2x + 3}}{{3 - 2x}}} \right)} \right\}

) is correct because it follows from applying the chain rule to differentiate the composite function.

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76 Easy

76 of 932

Application +4 | -1

Let

f(x) = x + \sin x,

suppose

g

denotes the inverse function of

f.

Then, find the value of

g'\left( {\frac{\pi }{4} + \frac{1}{{\sqrt 2 }}} \right)

View answer and explanation

Correct option: C

2 - \sqrt 2

Option C (

2 - \sqrt 2

) is correct because it is the value of the derivative of the inverse function g at the given point, calculated using the derivative of f.

77 Easy

77 of 932

Application +4 | -1

Let

f: R \rightarrow R

be a function such that

f(x)=x^3+x^2 f^{\prime}(1)+x f^{\prime \prime}(2)+f^{\prime \prime \prime}(3), x \in R

. Then,

f(2)

equals

View answer and explanation

Correct option: B

-2

Option B (-2) is correct because it is the value of the function f(x) at x=2, after determining the coefficients of the polynomial.

78 Medium

78 of 932

Application +4 | -1

f(x)=\sqrt{x}(x \geq 0)

and

g(x)=1+x^2

, then

(f \circ g)^{\prime}(1)

is equal to

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because, after finding

(f \circ g)(x) = \sqrt{1+x^2}

, we differentiate and evaluate at

x=1

, resulting in

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

79 Easy

79 of 932

Application +4 | -1

y=\log _2\left(\log _3\left(\log _5 x\right)\right)

then

y^{\prime}(125)

is equal to

View answer and explanation

Correct option: D

\frac{1}{375 \log 2 \log 3 \log 5}

Option D (

\frac{1}{375 \log 2 \log 3 \log 5}

) is correct because it is the result of correctly differentiating the logarithmic function and substituting x=125.

80 Easy

80 of 932

Application +4 | -1

x = a\left\{ {\mathrm{cos\theta } + \log \tan \left( {\frac{\mathrm{\theta }}{2}} \right)} \right\}

and

y = a\,\mathrm{sin\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 finding

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

by dividing

\frac{{dy}}{{d\theta}}

\frac{{dx}}{{d\theta}}

81 Medium

81 of 932

Application +4 | -1

f(x)=\sin x, g(x)=x^{2}, h(x)=\log x

and

F(x)=(

hogof)

(x)

, then

F^{\prime \prime}(x)

View answer and explanation

Correct option: D

-2 \operatorname{cosec}^{2} x

The correct answer is D (

-2 \operatorname{cosec}^{2} x

) because it is the second derivative of the function F(x) after simplification and differentiation.

82 Easy

82 of 932

Application +4 | -1

t=\sin ^{-1} 2^s

then

\frac{d s}{d t}

is equal to

View answer and explanation

Correct option: C

\frac{\cot t}{\log 2}

Option C (

\frac{{\cot t}}{{\log 2}}

) is correct because after finding ds/dt using implicit differentiation, it simplifies to the given expression.

83 Medium

83 of 932

Application +4 | -1

u\left( {x,y} \right) = y\log x + x\log y,

then

{u_x}{u_y} - {u_x}\log x - {u_y}\log y + \log x\log y

is equal to

View answer and explanation

Correct option: C

Option C (1) is correct because after calculating the partial derivatives and substituting them into the given expression, the expression simplifies to 1.

84 Medium

84 of 932

Application +4 | -1

Differential coefficient of

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

with respect to

{\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}}

will be

View answer and explanation

Correct option: A

Option A (1) is correct because the differential coefficient simplifies to 1 after trigonometric substitution and differentiation.

85 Easy

85 of 932

Application +4 | -1

Let

f(x)

be a polynomial in

x

. Then the second order derivative of

f\left(e^x\right)

with respect to

x

, is

View answer and explanation

Correct option: D

f^{\prime \prime}\left(e^x\right) e^{2 x}+f^{\prime}\left(e^x\right) \cdot e^x

Option D (

f^{\prime \prime}(e^x) e^{2 x}+f^{\prime}(e^x) \cdot e^x

) is correct because it is the result of applying the chain rule twice to find the second derivative.

86 Easy

86 of 932

Analyzing +4 | -1

u = - f''\,\,\left( \mathrm{\theta } \right)\,\,\sin \,\,\mathrm{\theta }\,\, + \,\,f'\,\,\left( \mathrm{\theta } \right)\,\,\cos \,\,\mathrm{\theta }

and

v = f''\,\,\left( \mathrm{\theta } \right)\,\,\cos \,\,\mathrm{\theta }\,\, + \,\,f'\,\,\left( \mathrm{\theta } \right)\,\,\sin \,\,\mathrm{\theta }

, then

f\,\,{\left[ {{{\left( {\frac{{du}}{{d\mathrm{\theta }}}} \right)}^2} + \,\,{{\left( {\frac{{dv}}{{d\mathrm{\theta }}}} \right)}^2}} \right]^{\frac{1}{2}}}d\mathrm{\theta }

is equal to

View answer and explanation

Correct option: B

f\,\,\left( \mathrm{\theta } \right)\,\, + \,\,f''\,\,\left( \mathrm{\theta } \right)\,\, + \,\,c

Option B (

f\,\left( \mathrm{\theta } \right)\, + \,f''\,\left( \mathrm{\theta } \right)\, + \,c

) is correct because after differentiating and simplifying, the integral evaluates to

f(\theta) + f''(\theta) + c

87 Medium

87 of 932

Application +4 | -1

y \sec x+\tan x+x^{2} y=0

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: C

-\frac{2 x y+\sec ^{2} x+y \sec x \tan x}{x^{2}+\sec x}

Option C (

-\frac{2 x y+\sec ^{2} x+y \sec x \tan x}{x^{2}+\sec x}

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

88 Easy

88 of 932

Application +4 | -1

f^{\prime}(x)=\sin (\log x)

and

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

, then

\frac{d y}{d x}

equals

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 (

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

) is correct because it is the derivative of the composite function calculated using the chain rule.

89 Medium

89 of 932

Application +4 | -1

y=\sin \left(2 \sin ^{-1} x\right)

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: A

\frac{2-4 x^{2}}{\sqrt{1-x^{2}}}

Option A is correct because it is the result of trigonometric substitution and differentiation.

90 Easy

90 of 932

Analyzing +4 | -1

y = \left[ {{{\tan }^{ - 1}}\frac{1}{{1 + x + {x^2}}} + {{\tan }^{ - 1}}\frac{1}{{{x^2} + 3x + 3}} + {{\tan }^{ - 1}}\frac{1}{{{x^2} + 5x + 7}} + ... + \,} \right.

upto

n

terms], they

y'\left( 0 \right)

is equal to

View answer and explanation

Correct option: C

\frac{{{n^2}}}{{\left( {{n^2} + 1} \right)}}

The correct answer is C (

\frac{{{n^2}}}{{\left( {{n^2} + 1} \right)}}

) because after simplifying the expression for y using the arctangent subtraction formula and differentiating, y'(0) evaluates to this value.

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