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 $y = {e^{{{\sin }^{ - 1}}x}}$ and $u = \log x$, then $\frac{{dy}}{{du}}$ is

$\frac{{{e^{{{\sin }^{ - 1}}x}}}}{{\sqrt {1 - {x^2}} }}$

$\frac{{{e^{{{\sin }^{ - 1}}x}}}}{x}$

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

$ \frac{-4}{\left(e^{2 x}-e^{-2 x}\right)^{2}} $

$ \frac{4}{\left(e^{2 x}-e^{-2 x}\right)^{2}} $

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JEE • MATHEMATICS • DIFFERENTIATION

Differentiation practice · Easy

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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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31 Easy

31 of 500

Analyzing +4 | -1

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

, then the value of

f\left( 0 \right) + f'\left( 0 \right) + \frac{{f''\left( 0 \right)}}{{2!}} + \ldots \frac{{{f^n}\left( 0 \right)}}{{n!}}

, is

View answer and explanation

Correct option: B

Option B (0) is correct because the given expression is the binomial expansion of

(1-1)^n

, which equals 0.

32 Easy

32 of 500

Application +4 | -1

The tangent line at

(2,4)

to the curve

y=

x^3-3 x+2

meets the

x

-axis at

View answer and explanation

Correct option: D

\left(\frac{14}{9}, 0\right)

Option D

\left(\frac{{14}}{{9}}, 0\right)

is correct because it represents the x-intercept of the tangent line to the given curve at the point (2,4).

33 Easy

33 of 500

Application +4 | -1

x = \sin t,y = \cos pt,

, then

View answer and explanation

Correct option: D

\left( {1 - {x^2}} \right){y_2} - x{y_1} + {p^2}y = 0

Option D (

\left( {1 - {x^2}} \right){y_2} - x{y_1} + {p^2}y = 0

\) is correct because it is the differential equation obtained after differentiating the given parametric equations and eliminating the parameter

t

34 Easy

34 of 500

Application +4 | -1

g

is the inverse of function

f

and

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

, then

g^{\prime}(x)=

View answer and explanation

Correct option: C

\frac{1}{\sin \,\{g(x)\}}

Option C (

\frac{1}{\sin \\{g(x)\\}}

) is correct because it is derived from differentiating the composite function

f(g(x)) = x

using the chain rule.

35 Easy

35 of 500

Application +4 | -1

The differential coefficient of

f\left(\log _{e} x\right)

w.r.t.

x

, where

f(x)=\log _{e} x

View answer and explanation

Correct option: C

\left(x \log _{e} x\right)^{-1}

Option C (

(x \log_e x)^{-1}

) is correct because it is the result of applying the chain rule to find the derivative of the composite function.

36 Easy

36 of 500

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.

37 Easy

37 of 500

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.

38 Easy

38 of 500

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.

39 Easy

39 of 500

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}

40 Easy

40 of 500

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).

41 Easy

41 of 500

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.

42 Easy

42 of 500

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.

43 Easy

43 of 500

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)

44 Easy

44 of 500

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}

45 Easy

45 of 500

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

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46 Easy

46 of 500

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.

47 Easy

47 of 500

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.

48 Easy

48 of 500

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.

49 Easy

49 of 500

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}}

50 Easy

50 of 500

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.

51 Easy

51 of 500

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.

52 Easy

52 of 500

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

53 Easy

53 of 500

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.

54 Easy

54 of 500

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.

55 Easy

55 of 500

Application +4 | -1

The derivative of

f\left( {\tan x} \right)\mathrm{w}\mathrm{.r}\mathrm{.t}\mathrm{.g}\left( {\sec x} \right)

x = \frac{\pi }{4},

where

f'\left( 1 \right) = 2

and

\mathrm{g'}\left( {\sqrt 2 } \right) = 4

, is

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because it is the result of differentiating f(tan x) with respect to g(sec x) and evaluating at x = pi/4.

56 Easy

56 of 500

Application +4 | -1

f(x)=\tan ^{-1}\left(\frac{3 \sin x}{4+5 \cos x}\right)

then

f^{\prime}\left(\frac{\pi}{3}\right)=

View answer and explanation

Correct option: A

\frac{3}{7}

Option A (

\frac{3}{7}

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

x = \frac{\pi}{3}

57 Easy

57 of 500

Application +4 | -1

\frac{{{d^2}}}{{d{x^2}}}(2\cos x\cos 3x)

is equal to

View answer and explanation

Correct option: D

- {2^2}(\cos 2x + {2^2}\cos 4x)

Option D (

- {2^2}(\cos 2x + {2^2}\cos 4x)

) is correct because it is the second derivative of the given function, calculated using trigonometric identities and differentiation.

58 Easy

58 of 500

Application +4 | -1

y = {e^{{{\sin }^{ - 1}}x}}

and

u = \log x

, then

\frac{{dy}}{{du}}

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because it is the result of finding

\frac{dy}{du}

using the chain rule.

59 Easy

59 of 500

Application +4 | -1

y=\frac{e^{2 x}+e^{-2 x}}{e^{2 x}-e^{-2 x}}

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: A

\frac{-8}{\left(e^{2 x}-e^{-2 x}\right)^{2}}

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

60 Easy

60 of 500

Application +4 | -1

x=a \cos 2 t, y=b \sin ^2 t

then

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

is equal to

View answer and explanation

Correct option: B

Option B (0) is correct because after finding the first derivative dy/dx, it results in a constant, and the derivative of a constant is zero.

Revision notes and key takeaways

Review DIFFERENTIATION in your textbook or class notes, then use this set to check your understanding.

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