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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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Showing 31-60 of 402 questions
31 Medium
31 of 402
Application +4 | -1
If f(x)=sin⁡x,g(x)=x2,h(x)=log⁡xf(x)=\sin x, g(x)=x^{2}, h(x)=\log x and F(x)=(F(x)=( hogof) (x)(x), then F′′(x)F^{\prime \prime}(x) is
View answer and explanation
Correct option: D
−2cosec⁡2x-2 \operatorname{cosec}^{2} x
The correct answer is D (−2cosec⁡2x-2 \operatorname{cosec}^{2} x) because it is the second derivative of the function F(x) after simplification and differentiation.
32 Medium
32 of 402
Application +4 | -1
If u(x,y)=ylog⁡x+xlog⁡y,u\left( {x,y} \right) = y\log x + x\log y, then uxuy−uxlog⁡x−uylog⁡y+log⁡xlog⁡y{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
1
Option C (1) is correct because after calculating the partial derivatives and substituting them into the given expression, the expression simplifies to 1.
33 Medium
33 of 402
Application +4 | -1
Differential coefficient of tan⁡−12x1−x2{\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}} with respect to sin⁡−12x1+x2{\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}} will be
View answer and explanation
Correct option: A
1
Option A (1) is correct because the differential coefficient simplifies to 1 after trigonometric substitution and differentiation.
34 Medium
34 of 402
Application +4 | -1
If ysec⁡x+tan⁡x+x2y=0y \sec x+\tan x+x^{2} y=0, then dydx\frac{d y}{d x} is equal to
View answer and explanation
Correct option: C
−2xy+sec⁡2x+ysec⁡xtan⁡xx2+sec⁡x-\frac{2 x y+\sec ^{2} x+y \sec x \tan x}{x^{2}+\sec x}
Option C (−2xy+sec⁡2x+ysec⁡xtan⁡xx2+sec⁡x-\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.
35 Medium
35 of 402
Application +4 | -1
If y=sin⁡(2sin⁡−1x)y=\sin \left(2 \sin ^{-1} x\right), then dydx\frac{d y}{d x} is equal to
View answer and explanation
Correct option: A
2−4x21−x2\frac{2-4 x^{2}}{\sqrt{1-x^{2}}}
Option A is correct because it is the result of trigonometric substitution and differentiation.
36 Medium
36 of 402
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.
37 Medium
37 of 402
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.
38 Medium
38 of 402
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.
39 Medium
39 of 402
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.
40 Medium
40 of 402
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.
41 Medium
41 of 402
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.
42 Medium
42 of 402
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.
43 Medium
43 of 402
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}.
44 Medium
44 of 402
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.
45 Medium
45 of 402
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.

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46 Medium
46 of 402
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).
47 Medium
47 of 402
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.
48 Medium
48 of 402
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.
49 Medium
49 of 402
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.
50 Medium
50 of 402
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}.
51 Medium
51 of 402
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.
52 Medium
52 of 402
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.
53 Medium
53 of 402
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.
54 Medium
54 of 402
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}.
55 Medium
55 of 402
Application +4 | -1
If x=Acos⁡4t+Bsin⁡4tx = A\cos 4t + B\sin 4t, then d2xdt2\frac{{{d^2}x}}{{d{t^2}}} is equal to
View answer and explanation
Correct option: A
−16x- 16x
Option A (-16x) is correct because it is the second derivative of the given function, which represents simple harmonic motion.
56 Medium
56 of 402
Application +4 | -1
A differentiable function f(x)f\left( x \right) is defined for all x>0x > 0 and satisfies f(x3)=4x4f\left( {{x^3}} \right) = 4{x^4} for all x>0.x > 0. The value of f′(8)f'\left( 8 \right) is
View answer and explanation
Correct option: B
323\frac{{32}}{3}
Option B (323\frac{{32}}{3}) is correct because it is the value of the derivative f′(8)f'(8) obtained by differentiating the given function f(x3)=4x4f(x^3) = 4x^4 and substituting x=2x=2.
57 Medium
57 of 402
Application +4 | -1
If x=exp⁡{tan⁡−1(y−x2x2)}x = \exp \left\{ {{{\tan }^{ - 1}}\left( {\frac{{y - {x^2}}}{{{x^2}}}} \right)} \right\}, then dydx\frac{{dy}}{{dx}} equal
View answer and explanation
Correct option: A
2x{1+tan⁡(log⁡ex)}+xsec⁡2(log⁡ex)2x\left\{ {1 + \tan \left( {{{\log }_e}x} \right)} \right\} + x{\sec ^2}\left( {{{\log }_e}x} \right)
Option A (2x{1+tan⁡(log⁡ex)}+xsec⁡2(log⁡ex)2x\{{1 + \tan ({{{\log }_e}x})} \} + x{\sec ^2}({{{\log }_e}x})) is correct because it is the result of correctly differentiating the given implicit function.
58 Medium
58 of 402
Application +4 | -1
If In=dndxn(xnlog⁡x),{I_n} = \frac{{{d^n}}}{{d{x^n}}}({x^n}\log x), then In−nln−1{I_n} - n{l_{n - 1}} is equal to
View answer and explanation
Correct option: D
(n−1)!\left( {n - 1} \right)!
Option D ((n−1)!(n - 1)!) is correct because the expression In−nln−1{I_n} - n{l_{n - 1}} simplifies to (n−1)!(n - 1)! after applying the given recurrence relation.
59 Medium
59 of 402
Analyzing +4 | -1
If f(x)=cos⁡xcos⁡2xcos⁡4xcos⁡8xcos⁡16xf\left( x \right) = \cos x\cos 2x\cos 4x\cos 8x\cos 16x, then f′(π4)f'\left( {\frac{\pi }{4}} \right) is
View answer and explanation
Correct option: C
0
The correct answer is C (0) because f(π4)=0f(\frac{\pi}{4}) = 0, which makes f′(π4)=0f'(\frac{\pi}{4}) = 0.
60 Medium
60 of 402
Application +4 | -1
If y=cosec⁡(cot⁡−1x)y=\operatorname{cosec}\left(\cot ^{-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 differentiating yy and substituting x=1x=1 yields the result 12\frac{{1}}{{\sqrt{{2}}}}.
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