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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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91 Easy
91 of 932
Application +4 | -1
The derivative of f(tan⁡x)w.r.t.g(sec⁡x)f\left( {\tan x} \right)\mathrm{w}\mathrm{.r}\mathrm{.t}\mathrm{.g}\left( {\sec x} \right) at x=π4,x = \frac{\pi }{4}, where f′(1)=2f'\left( 1 \right) = 2 and g′(2)=4\mathrm{g'}\left( {\sqrt 2 } \right) = 4, is
View answer and explanation
Correct option: A
12\frac{1}{{\sqrt 2 }}
Option A (12\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.
92 Medium
92 of 932
Application +4 | -1
If y=sin⁡(log⁡ex)y = \sin \left( {{{\log }_e}x} \right), then x2d2ydx2+xdydx{x^2}\frac{{{d^2}y}}{{d{x^2}}} + x\frac{{dy}}{{dx}} is equal to
View answer and explanation
Correct option: D
−y- y
Option D (−y- y) is correct because after differentiating the given function twice and rearranging, the expression x2d2ydx2+xdydx{x^2}\frac{{{d^2}y}}{{d{x^2}}} + x\frac{{dy}}{{dx}} simplifies to −y-y.
93 Medium
93 of 932
Application +4 | -1
If y=axn+1+bx−ny=a x^{n+1}+b x^{-n}, then x2d2ydx2=x^2 \frac{d^2 y}{d x^2}=
View answer and explanation
Correct option: B
n(n+1)yn(n+1) y
Option B (n(n+1)yn(n+1) y) is correct because it is the result of calculating the second derivative of y=axn+1+bx−ny=a x^{n+1}+b x^{-n} and then multiplying by x2x^2.
94 Medium
94 of 932
Analyzing +4 | -1
If x=acos⁡θ,y=bsin⁡θx = a\cos \theta ,y = b\sin \theta, then d3ydx3\frac{{{d^3}y}}{{d{x^3}}} is equal to
View answer and explanation
Correct option: C
−3ba3cosec4θcot⁡θ- \frac{{3b}}{{{a^3}}}\mathrm{cose}{\mathrm{c}^4}\theta \cot \theta
Option D (None of these) is correct. After calculating the derivatives, the correct expression for d3ydx3\frac{{{{d^3}y}}}{{{{dx^3}}}} does not match any of the given options.
95 Medium
95 of 932
Application +4 | -1
If x2ey+2xyex+13=0,{x^2}{e^y} + 2xy{e^x} + 13 = 0, then dydx\frac{{dy}}{{dx}} equals
View answer and explanation
Correct option: B
−[2xey−x+2y(x+1)x(xey−x+2)]- \left[ {\frac{{2x{e^{y - x}} + 2y(x + 1)}}{{x(x{e^{y - x}} + 2)}}} \right]
Option B (−[2xey−x+2y(x+1)x(xey−x+2)]- \left[ {\frac{{2x{e^{y - x}} + 2y(x + 1)}}{{x(x{e^{y - x}} + 2)}}} \right]\) is correct because it represents the derivative dy/dx obtained through implicit differentiation of the given equation.
96 Medium
96 of 932
Analyzing +4 | -1
If f(x)=arctan⁡(xx−x−x2)f\left( x \right) = \mathrm{arc}\tan \left( {\frac{{{x^x} - {x^{ - x}}}}{2}} \right), then f′(1)f'\left( 1 \right) is equal to
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the derivative f′(1)f'(1) does not exist due to the function's behavior around x=1x=1.
97 Easy
97 of 932
Application +4 | -1
If f(x)=tan⁡−1(3sin⁡x4+5cos⁡x)f(x)=\tan ^{-1}\left(\frac{3 \sin x}{4+5 \cos x}\right) then f′(π3)=f^{\prime}\left(\frac{\pi}{3}\right)=
View answer and explanation
Correct option: A
37\frac{3}{7}
Option A (37\frac{3}{7}) is correct because it is the result of differentiating the given function and substituting x=π3x = \frac{\pi}{3}.
98 Medium
98 of 932
Application +4 | -1
If x=asin⁡−1tx=\sqrt{a^{\sin ^{-1} t}} and y=acos⁡−1ty=\sqrt{a^{\cos ^{-1} t}} then dydx=\frac{d y}{d x}=
View answer and explanation
Correct option: D
−yx-\frac{y}{x}
Option D (−yx\frac{-y}{x}) is correct because multiplying the given equations and differentiating leads to this result.
99 Easy
99 of 932
Application +4 | -1
d2dx2(2cos⁡xcos⁡3x)\frac{{{d^2}}}{{d{x^2}}}(2\cos x\cos 3x) is equal to
View answer and explanation
Correct option: D
−22(cos⁡2x+22cos⁡4x)- {2^2}(\cos 2x + {2^2}\cos 4x)
Option D (−22(cos⁡2x+22cos⁡4x)- {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.
100 Easy
100 of 932
Application +4 | -1
If y=esin⁡−1xy = {e^{{{\sin }^{ - 1}}x}} and u=log⁡xu = \log x, then dydu\frac{{dy}}{{du}} is
View answer and explanation
Correct option: C
xesin⁡−1x1−x2\frac{{x{e^{{{\sin }^{ - 1}}x}}}}{{\sqrt {1 - {x^2}} }}
Option C (xesin⁡−1x1−x2\frac{{x{e^{{{\sin }^{ - 1}}x}}}}{{\sqrt {1 - {x^2}} }}) is correct because it is the result of finding dydu\frac{dy}{du} using the chain rule.
101 Medium
101 of 932
Application +4 | -1
If y3d2ydx2+k=0y^3 \frac{d^2 y}{d x^2}+k=0 for y=2x−x2y=\sqrt{2 x-x^2}, then k=……k=\ldots \ldots
View answer and explanation
Correct option: A
1
Option A (1) is correct because after finding the first and second derivatives and substituting them into the given equation, the value of k is found to be 1.
102 Easy
102 of 932
Application +4 | -1
If y=e2x+e−2xe2x−e−2xy=\frac{e^{2 x}+e^{-2 x}}{e^{2 x}-e^{-2 x}}, then dydx\frac{d y}{d x} is equal to
View answer and explanation
Correct option: A
−8(e2x−e−2x)2\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.
103 Easy
103 of 932
Application +4 | -1
If x=acos⁡2t,y=bsin⁡2tx=a \cos 2 t, y=b \sin ^2 t then d2ydx2\frac{d^2 y}{d x^2} is equal to
View answer and explanation
Correct option: B
0
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.
104 Easy
104 of 932
Application +4 | -1
The derivative of y=(1−x)(2−x)…(n−x)y = \left( {1 - x} \right)\left( {2 - x} \right) \ldots \left( {n - x} \right) at x=1x = 1 is equal to
View answer and explanation
Correct option: B
(−1)(n−1)!\left( { - 1} \right)\left( {n - 1} \right)!
Option B ((−1)(n−1)!\left( { - 1} \right)\left( {n - 1} \right)!\) is correct because it is the value of the derivative of the given product at x=1x = 1, obtained by using logarithmic differentiation.
105 Easy
105 of 932
Application +4 | -1
If y=log⁡sin⁡x(tan⁡x)y=\log _{\sin x}(\tan x), then dydx\frac{d y}{d x} at x=π4x=\frac{\pi}{4} is equal to
View answer and explanation
Correct option: C
−4log⁡2\frac{-4}{\log 2}
Option C (−4log⁡2\frac{-4}{\log 2}) is correct because it is the result of differentiating the logarithmic function and evaluating at x=π4x=\frac{\pi}{4}.

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs
106 Medium
106 of 932
Analyzing +4 | -1
If x=etsin⁡t,y=etcos⁡t,tx = {e^t}\sin t,y = {e^t}\cos t,t is a parameter, then d2ydx2\frac{{{d^2}y}}{{d{x^2}}} at (1,1) is equal to
View answer and explanation
Correct option: A
−12- \frac{1}{2}
The correct answer is A (−12- \frac{1}{2}) because after finding the first and second derivatives using parametric differentiation and substituting t=π4t = \frac{\pi}{4}, the value of d2ydx2\frac{d^2y}{dx^2} at (1,1) is −12-\frac{1}{2}.
107 Easy
107 of 932
Application +4 | -1
If y=xx2,y = {x^{{x^2}}}, then dydx\frac{{dy}}{{dx}} equals
View answer and explanation
Correct option: D
xx2+1.  (log⁡(ex2)){x^{{x^2} + 1}}.\,\,(\log (e{x^2}))
Option D (xx2+1.(log⁡(ex2)){x^{{x^2} + 1}}.{\left( {\log (e{x^2})} \right)}) is correct because it is the derivative of the function y = x(x2) calculated using logarithmic differentiation.
108 Easy
108 of 932
Application +4 | -1
If xm⋅yn=(x+y)m+nx^{m} \cdot y^{n}=(x+y)^{m+n}, then dydx\frac{d y}{d x} is
View answer and explanation
Correct option: A
yx\frac{y}{x}
Option A (yx\frac{y}{x}) is correct because after taking the logarithm and differentiating the given equation, the expression simplifies to dydx=yx\frac{dy}{dx} = \frac{y}{x}.
109 Medium
109 of 932
Analyzing +4 | -1
If x2+y2=R2(R>0),{x^2} + {y^2} = {R^2}(R > 0), then K=y′′1+(y′)2)3.K = \frac{{y''}}{{\sqrt {1 + {{(y')}^2}{)^3}} }}. Find KK in terms of R.R.
View answer and explanation
Correct option: B
−1R- \frac{1}{R}
Option B (−1R- \frac{1}{R}) is correct because it is the curvature K of the circle expressed in terms of its radius R.
110 Medium
110 of 932
Application +4 | -1
A point moves according s=29sin⁡π2t+s0s = \frac{2}{9}\sin \frac{\pi }{2}t + {s_0} . The acceleration at the end of first second is
View answer and explanation
Correct option: B
−π218- \frac{{{\pi ^2}}}{{18}}
Option B (−π218- \frac{{{\pi ^2}}}{{18}}) is correct because it is the acceleration calculated by taking the second derivative of the position function and evaluating at t=1.
111 Medium
111 of 932
Application +4 | -1
If f(x)=beax+aebxf\left( x \right) = b{e^{ax}} + a{e^{bx}}, then f′′(0)f''\left( 0 \right) is equal to
View answer and explanation
Correct option: C
ab(a+b)ab\left( {a + b} \right)
Option C (ab(a + b)) is correct because finding the second derivative and evaluating at x=0 gives ab(a+b).
112 Medium
112 of 932
Application +4 | -1
If 2x2−3xy+y2+x+2y−x=02{x^2} - 3xy + {y^2} + x + 2y - x = 0, then dydx=\frac{{dy}}{{dx}} =
View answer and explanation
Correct option: A
3y−4x−12y−3x+2\frac{{3y - 4x - 1}}{{2y - 3x + 2}}
Option A (3y−4x−12y−3x+2\frac{{3y - 4x - 1}}{{2y - 3x + 2}}) is correct because it is the result of implicit differentiation of the given equation with respect to x.
113 Medium
113 of 932
Analyzing +4 | -1
If f(x)=x+cos⁡xx−cos⁡xf(x)=\frac{x+\cos x}{x-\cos x}, then lim⁡x→0f(x)−f(0)x\lim \limits_{{x \rightarrow 0}} \frac{f(x)-f(0)}{x} is
View answer and explanation
Correct option: A
−2-2
Option A (-2) is correct because the limit represents the derivative of f(x) at x=0, and calculating f'(0) yields -2.
114 Medium
114 of 932
Understanding +4 | -1
Let f(x)f(x) be a continous function on RR such that f(14n)=sin⁡(en)e−n2+n2n2+1f\left(\frac{1}{4 n}\right)=\sin \left(e^n\right) e^{-n^2}+\frac{n^2}{n^2+1}. Then the value of f(0)f(0) is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the limit of the given function as n approaches infinity equals 1, and since f(x) is continuous, f(0) equals that limit.
115 Medium
115 of 932
Analyzing +4 | -1
For x>1\mathrm{x}>1, if (2x)2y=4e2x−2y(2 x)^{2 y}=4 e^{2 x-2 y}, then (1+log⁡e2x)2dydx\left(1+\log _{e} 2 x\right)^{2} \frac{d y}{d x} is equal to :
View answer and explanation
Correct option: A
xlog⁡e2x−log⁡e2x\frac{x \log _{e} 2 x-\log _{e} 2}{x}
Option A (xlog⁡e2x−log⁡e2x\frac{x \log _{e} 2 x-\log _{e} 2}{x}) is correct because it is the result of differentiating the given equation and solving for the expression (1+log⁡e2x)2dydx(1+\log _{e} 2 x)^{2} \frac{d y}{d x}.
116 Easy
116 of 932
Application +4 | -1
If y=sec⁡(tan⁡−1x)y=\sec \left(\tan ^{-1} x\right), then dydx\frac{d y}{d x} at x=1x=1 is equal to
View answer and explanation
Correct option: A
12\frac{1}{\sqrt{2}}
Option A (12\frac{1}{\sqrt{2}}) is correct because it is the value of the derivative of the given function at the specified point.
117 Medium
117 of 932
Application +4 | -1
y=easin⁡−1x⇒(1−x2)yn+2−(2n+1)xyn+1y = {e^{a{{\sin }^{ - 1}}x}} \Rightarrow \left( {1 - {x^2}} \right){y_{n + 2}} - \left( {2n + 1} \right)x{y_{n + 1}} is equal to
View answer and explanation
Correct option: C
(n2+a2)yn\left( {{n^2} + {a^2}} \right){y_n}
Option C ((n2+a2)yn\left( {{n^2} + {a^2}} \right){y_n}) is correct because it is the correct result obtained by applying Leibnitz's rule to the given differential equation.
118 Medium
118 of 932
Analyzing +4 | -1
Let f(x)={(x−1)sin⁡1x−1 if x≠10 if x=1f(x)=\left\{\begin{array}{cc}(x-1) \sin \frac{1}{x-1} \text { if } x \neq 1 \\ 0 & \text { if } x=1\end{array}\right.
Then which one of the following is true?
View answer and explanation
Correct option: C
ff is differentiable at x=0x=0 but not at x=1x=1
Option C is correct because f(x) is differentiable at x=0 but not at x=1, as the left-hand and right-hand derivatives are not equal at x=1.
119 Medium
119 of 932
Application +4 | -1
If y=f(2x−1x2+1)y=f\left(\frac{2 x-1}{x^2+1}\right) and f′(x)=sin⁡x2f^{\prime}(x)=\sin x^2, then dydx\frac{d y}{d x} is equal to
View answer and explanation
Correct option: B
sin⁡(2x−1x2+1)2{2+2x−2x2(x2+1)2}\sin \left(\frac{2 x-1}{x^2+1}\right)^2\left\{\frac{2+2 x-2 x^2}{\left(x^2+1\right)^2}\right\}
Option B is correct because it correctly applies the chain rule and quotient rule to find the derivative of the given function.
120 Medium
120 of 932
Application +4 | -1
If x=1−t21+t2x = \frac{{1 - {t^2}}}{{1 + {t^2}}} and y=2t1+t2,y = \frac{{2t}}{{1 + {t^2}}}, then dydx\frac{{dy}}{{dx}} is equal to
View answer and explanation
Correct option: C
−xy- \frac{x}{y}
Option C (−xy\frac{{ - x}}{y}) is correct because after substituting t=tan⁡θt = \tan \theta and differentiating, the result is −xy\frac{{ - x}}{y}.
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