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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 421-450 of 932 questions

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421 Medium
421 of 932
Analyzing +4 | -1
If x2+y2=t−1t{x^2} + {y^2} = t - \frac{1}{t} and x4+y4=t2+1t2,{x^4} + {y^4} = {t^2} + \frac{1}{{{t^2}}}, then dydx\frac{{dy}}{{dx}} is equal to
View answer and explanation
Correct option: D
1x3y\frac{1}{{{x^3}y}}
The correct answer is D (1x3y\frac{{1}}{{{{x^3}y}}}) because it is obtained by differentiating the given equations and solving for dydx\frac{{dy}}{{dx}}.
422 Easy
422 of 932
Application +4 | -1
If f′(x)=tan⁡−1(sec⁡ x+tan⁡ x), π2<x<π2f'\left( x \right) = {\tan ^{ - 1}}\left( {\sec \,x + \tan \,x} \right),\,\frac{\pi }{2} < x < \frac{\pi }{2} and f(0)=0,f\left( 0 \right) = 0, then f(1)f\left( 1 \right) is equal to
View answer and explanation
Correct option: A
π+14\frac{{\pi + 1}}{4}
Option A ((π\pi + 1)/4) is correct because, after simplifying f'(x) and integrating, using the initial condition f(0) = 0, we find f(1) = (π\pi + 1)/4.
423 Medium
423 of 932
Application +4 | -1
The derivative of f(x)=log⁡x23f(x)=\log _{x^2} 3, w. r. to xx is.
View answer and explanation
Correct option: C
−ln⁡32x(ln⁡x)2-\frac{\ln 3}{2 x(\ln x)^2}
Option C (−ln⁡32x(ln⁡x)2-\frac{\ln 3}{2 x(\ln x)^2}) is correct because it is the result of applying the chain rule and change of base formula to differentiate the given logarithmic function.
424 Medium
424 of 932
Application +4 | -1
d2xdy2\frac{d^2 x}{d y^2} equals
View answer and explanation
Correct option: D
−(d2ydx2)(dydx)−3-\left(\frac{d^2 y}{d x^2}\right)\left(\frac{d y}{d x}\right)^{-3}
Option D is correct because it correctly applies the chain rule and the derivative of an inverse function.
425 Easy
425 of 932
Application +4 | -1
If x=ey+ey+⋯ too ,x>0x=e^{y+e^{y+\cdots \text { too }}}, x>0, then dydx\frac{d y}{d x} is
View answer and explanation
Correct option: C
1−xx\frac{1-x}{x}
Option C (1−xx\frac{{1-x}}{{x}}) is correct because the derivative dydx\frac{{dy}}{{dx}} is found by implicit differentiation after taking the logarithm of both sides of the given equation.
426 Easy
426 of 932
Application +4 | -1
If f(x)=xtan⁡−1xf(x)=x \tan ^{-1} x, then f′(1)f^{\prime}(1) equals
View answer and explanation
Correct option: A
12+π4\frac{1}{2}+\frac{\pi}{4}
Option A (12+π4\frac{1}{2}+\frac{\pi}{4}) is correct because it is the value of the derivative of f(x)=xtan⁡−1xf(x)=x \tan ^{-1} x evaluated at x=1x=1.
427 Easy
427 of 932
Analyzing +4 | -1
If y=(1+1x)(1+2x)(1+3x)…(1+nx)y = \left( {1 + \frac{1}{x}} \right)\left( {1 + \frac{2}{x}} \right)\left( {1 + \frac{3}{x}} \right) \ldots \left( {1 + \frac{n}{x}} \right) and x≠0,x \ne 0, then dydx\frac{{dy}}{{dx}} when x=−1x = - 1 is
View answer and explanation
Correct option: C
(−1)n(n−1)!{\left( { - 1} \right)^n}\left( {n - 1} \right)!
Option D ((−1)n(n−1)!{\left( { - 1} \right)^n}\left( {n - 1} \right)!) is correct because it is obtained by differentiating the given product and substituting x=−1x = -1.
428 Hard
428 of 932
Application +4 | -1
x=etan⁡−1(y−x2x2)x=e^{\tan ^{-1}\left(\frac{y-x^{2}}{x^{2}}\right)}, then dydx\frac{d y}{d x} is
View answer and explanation
Correct option: A
x(2+y2x4−2yx2)+2yxx\left(2+\frac{y^{2}}{x^{4}}-\frac{2 y}{x^{2}}\right)+\frac{2 y}{x}
The correct answer is A (x(2+y2x4−2yx2)+2yxx\left(2+\frac{y^{2}}{x^{4}}-\frac{2 y}{x^{2}}\right)+\frac{2 y}{x}) because it is the derivative of y with respect to x after implicit differentiation and simplification.
429 Medium
429 of 932
Application +4 | -1
The derivative of sin⁡2x{\sin ^2}x with respect to cos⁡2x{\cos ^2}x is
View answer and explanation
Correct option: D
None of these
The correct answer is None of these because the derivative of sin²x with respect to cos²x is -1, which is not among the given options.
430 Easy
430 of 932
Analyzing +4 | -1
If 8f(x)+6f(1x)=x+58f\left( x \right) + 6f\left( {\frac{1}{x}} \right) = x + 5 and y=x2f(x)y = {x^2}f\left( x \right), then the value of dydx\frac{{dy}}{{dx}} at x=−1x = - 1, is
View answer and explanation
Correct option: C
−114- \frac{1}{{14}}
Option C ( −114\ - \frac{1}{{14}}) is correct because after solving for f(x)f(x) and substituting it into y=x2f(x)y = x^2f(x), the derivative at x=−1x = -1 is −114-\frac{1}{14}.
431 Easy
431 of 932
Application +4 | -1
If f(x)=x+2f\left( x \right) = x + 2, then f′(f(x))f'\left( {f\left( x \right)} \right) at x=4x = 4, is
View answer and explanation
Correct option: B
1
Option B (1) is correct because the derivative of f(x) = x+2 is always 1, regardless of the input.
432 Easy
432 of 932
Application +4 | -1
If f(x)=xtan⁡−1xf(x)=x \tan ^{-1} x, then f′(1)f^{\prime}(1) is equal to
View answer and explanation
Correct option: A
12+π4\frac{1}{2}+\frac{\pi}{4}
Option A (12+π4\frac{{1}}{{2}}+\frac{{\pi}}{{4}}) is correct because it is the value of the derivative of f(x)=xtan⁡−1xf(x)=x \tan^{-1}x at x=1x=1.
433 Medium
433 of 932
Analyzing +4 | -1
Let f(x)=−4⋅e2−x4+10+2x+x3+x5f(x)=-4 \cdot e^{\frac{2-x}{4}}+10+2 x+x^3+x^5 and g(x)=f−1(x)g(x)=f^{-1}(x), then the value of g′(50)g^{\prime}(50) is
View answer and explanation
Correct option: C
195\frac{1}{95}
Option C (195\frac{1}{95}) is correct because using the inverse function derivative formula g′(x)=1f′(g(x))g'(x) = \frac{1}{f'(g(x))}, we find g′(50)=1f′(2)=195g'(50) = \frac{1}{f'(2)} = \frac{1}{95}.
434 Medium
434 of 932
Analyzing +4 | -1
Let (x)=log⁡f(x),\left( x \right) = \log f\left( x \right), where f(x)f\left( x \right) is a twice differentiable positive function on (0,∞)\left( {0,\infty } \right) such that f(x+1)=xf(x).f\left( {x + 1} \right) = xf\left( x \right). Then, for N=1,2,3,…,g′′(N+12)−g′′(12)N = 1,2,3, \ldots ,\mathrm{g''}\left( {N + \frac{1}{2}} \right) - \mathrm{g''}\left( {\frac{1}{2}} \right) is equal to
View answer and explanation
Correct option: A
−4{1+19+125+…+1(2N−1)2}- 4\left\{ {1 + \frac{1}{9} + \frac{1}{{25}} + \ldots + \frac{1}{{{{\left( {2N - 1} \right)}^2}}}} \right\}
Option A (−4{1+19+125+…+1(2N−1)2}- 4\left\{ {1 + \frac{1}{9} + \frac{1}{{25}} + \ldots + \frac{1}{{{{\left( {2N - 1} \right)}^2}}}} \right\}) is correct because it is the result obtained after substituting and simplifying the given expression.
435 Easy
435 of 932
Analyzing +4 | -1
Let f:R→Rf: R \rightarrow R be a function. We say that ff has

Property 1: If lim⁡h→0f(h)−f(0)∣h∣\lim\limits _{h \rightarrow 0} \frac{f(h)-f(0)}{\sqrt{|h|}} exists and is finite.
and,
Property 2: If lim⁡h→0f(h)−f(0)h2\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h^2} exists and is finite.

Then which of the following option is correct?
View answer and explanation
Correct option: B
f(x)=x2/3f(x)=x^{2 / 3} has Property 1
Option B is correct because the limit lim⁡h→0f(h)−f(0)∣h∣\lim_{h \rightarrow 0} \frac{f(h)-f(0)}{\sqrt{|h|}} exists and is finite for f(x)=x2/3f(x)=x^{2 / 3}.

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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
436 Easy
436 of 932
Application +4 | -1
If x=acos⁡3θ,y=asin⁡3θx=a \cos ^3 \theta, y=a \sin ^3 \theta, then 1+(dydx)2=\sqrt{1+\left(\frac{d y}{d x}\right)^2}=
View answer and explanation
Correct option: D
∣sec⁡θ∣|\sec \theta|
Option D (∣sec⁡θ∣|\sec \theta|) is correct because differentiating and simplifying the expression leads to the absolute value of secant.
437 Medium
437 of 932
Application +4 | -1
ddx(sec⁡x+tan⁡xsec⁡x−tan⁡x)\frac{d}{d x}\left(\frac{\sec x+\tan x}{\sec x-\tan x}\right) is equal to
View answer and explanation
Correct option: A
2cos⁡x(1−sin⁡x)2\frac{2 \cos x}{(1-\sin x)^{2}}
Option A is correct because the expression is simplified using trigonometric identities and then differentiated.
438 Medium
438 of 932
Analyzing +4 | -1
Let g(x)=(x−1)nlog⁡cos⁡m(x−1),0<x<2,mg\left( x \right) = \frac{{{{\left( {x - 1} \right)}^n}}}{{\log {{\cos }^m}\left( {x - 1} \right)}},0 < x < 2,m and nn are integers, m≠0,n>0m \ne 0,n > 0, and let pp be the left hand derivative of ∣x−1∣\left| {x - 1} \right| at x=1x = 1. If lim⁡x→1+g(x)=p\lim\limits_{x \to {1^ + }} g\left( x \right) = p, then
View answer and explanation
Correct option: C
n=2,m=2n = 2,m = 2
Option C (n=2,m=2n = 2,m = 2) is correct because it satisfies the limit condition lim⁡x→1+g(x)=p=−1\lim\limits_{{x \to {1^ + }}} g(x) = p = -1.
439 Medium
439 of 932
Application +4 | -1
If 2y=(cot⁡−1(3cos⁡x+sin⁡xcos⁡x−3sin⁡x))2,x∈(0,π2)2 y=\left(\cot ^{-1}\left(\frac{\sqrt{3} \cos x+\sin x}{\cos x-\sqrt{3} \sin x}\right)\right)^{2}, x \in\left(0, \frac{\pi}{2}\right) then dydx\frac{d y}{d x} is equal to :
View answer and explanation
Correct option: A
π6−x\frac{\pi}{6}-x
Option A (π6−x\frac{\pi}{6}-x) is correct because by simplifying the expression inside the cotangent inverse function, we get 2y=(π6−x)22y = (\frac{\pi}{6} - x)^2, and differentiating this gives dydx=π6−x\frac{dy}{dx} = \frac{\pi}{6}-x.
440 Medium
440 of 932
Analyzing +4 | -1
If f(x)=(log⁡cot⁡xlog⁡cot⁡xtan⁡x)(log⁡tan⁡xlog⁡tan⁡xcot⁡x)−1+tan⁡−14x4−x2,f\left( x \right) = ({\log _{\cot x}}{\log _{\cot x}}\tan x)({\log _{\tan x}}{\log _{\tan x}}\cot {\left. x \right)^{ - 1}} + {\tan ^{ - 1}}\frac{{4x}}{{4 - {x^2}}}, then f′(2)f'\left( 2 \right) is equal to
View answer and explanation
Correct option: A
12\frac{1}{2}
Option A (12\frac{1}{2}) is correct because after simplifying the function and differentiating, the value of the derivative at x=2 is 12\frac{1}{2}.
441 Easy
441 of 932
Application +4 | -1
ddx[sin⁡−1(x1−x−x1−x2)]\frac{d}{{dx}}\left[ {{{\sin }^{ - 1}}(x\sqrt {1 - x} - \sqrt x \sqrt {1 - {x^2})} } \right] is equal to
View answer and explanation
Correct option: C
11−x2−12x(1−x)\frac{1}{{\sqrt {1 - {x^2}} }} - \frac{1}{{2\sqrt {x\left( {1 - x} \right)} }}
Option C (11−x2−12x(1−x)\frac{1}{{\sqrt {1 - {x^2}} }} - \frac{1}{{2\sqrt {x\left( {1 - x} \right)} }}) is correct because it represents the derivative of the given expression after applying trigonometric substitution and simplification.
442 Medium
442 of 932
Application +4 | -1
If f(x)=(x+1)tan⁡−1(e−2x)f\left( x \right) = \left( {x + 1} \right){\tan ^{ - 1}}\left( {{e^{ - 2x}}} \right), then f′(0)f'\left( 0 \right) is
View answer and explanation
Correct option: B
π4−1\frac{\pi }{4} - 1
Option B (π4−1\frac{\pi }{4} - 1) is correct because after differentiating and substituting x=0x = 0, the value of f′(0)f'(0) is π4−1\frac{\pi }{4} - 1.
443 Medium
443 of 932
Analyzing +4 | -1
Let f(x+y)=f(x)+f(y)f\left( {x + y} \right) = f\left( x \right) + f\left( y \right) and f(x)=x2g(x)f\left( x \right) = {x^2}g\left( x \right) for all x,y∈Rx,y \in R, where g(x)g\left( x \right) is continuous function. Then, f′(x)f'\left( x \right) is equal to
View answer and explanation
Correct option: D
0
Option D (0) is correct because, given f(x+y) = f(x) + f(y) and f(x) = x²g(x), it implies f(x) = 0 for all x, thus f'(x) = 0.
444 Hard
444 of 932
Analyzing +4 | -1
Let f(x+y)=f(x)f(y)f\left( {x + y} \right) = f\left( x \right)f\left( y \right) and f(x)=1+xg(x)G(x)f\left( x \right) = 1 + xg\left( x \right)G\left( x \right), where lim⁡x→0g(x)=a\lim\limits_{x \to 0} g\left( x \right) = a and lim⁡x→0G(x)=b\lim\limits_{x \to 0} G\left( x \right) = b. Then f′(x)f'\left( x \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′(x)f'(x) is equal to abf(x)ab f(x), which is not listed in the options.
445 Medium
445 of 932
Analyzing +4 | -1
If F(x)=1x2∫4x(4t2−2F′(t))dtF\left( x \right) = \frac{1}{{{x^2}}}\int \limits_4^x \left( {4{t^2} - 2F'\left( t \right)} \right)dt , then F′(4)F'\left( 4 \right) equals
View answer and explanation
Correct option: A
329\frac{{32}}{9}
Option A (329\frac{{32}}{9}) is correct because after differentiating both sides and substituting x=4, we find that F'(4) = 329\frac{{32}}{9}.
446 Hard
446 of 932
Analyzing +4 | -1
The following functions are differentiable on (−1,2)\left( { - 1,2} \right)
View answer and explanation
Correct option: C
∫x2x1−t+t21+t+t2dt\int \limits_x^{2x} \frac{{1 - t + {t^2}}}{{1 + t + {t^2}}}dt
Option C (∫x2x1−t+t21+t+t2dt\int \limits_x^{2x} \frac{{1 - t + {t^2}}}{{1 + t + {t^2}}}dt) is correct because the integrand is continuous on the interval (-1, 2), ensuring differentiability of the integral.
447 Medium
447 of 932
Analyzing +4 | -1
If x=2sin⁡θ−sin⁡2θx=2 \sin \theta-\sin 2 \theta and y=2cos⁡θ−cos⁡2θ,θ∈[0,2π]y=2 \cos \theta-\cos 2 \theta, \theta \in[0,2 \pi], then d2ydx2\frac{d^{2} y}{d x^{2}} at θ=π\theta=\pi is :
View answer and explanation
Correct option: B
38\frac{3}{8}
Option B (38\frac{3}{8}) is correct because it is the value of the second derivative of y with respect to x, evaluated at θ=π\theta = \pi.
448 Easy
448 of 932
Application +4 | -1
Consider f(x)=tan⁡−1(1+sin⁡x1−sin⁡x),  x∈(0,π2)f\left( x \right) = {\tan ^{ - 1}}\left( {\sqrt {\frac{{1 + \sin x}}{{1 - \sin x}}} } \right),\,\,x \in \left( {0,\frac{\pi }{2}} \right).
A normal to y=f(x)y = f\left( x \right) at x=π6x = \frac{\pi }{6} also passes through the point
View answer and explanation
Correct option: B
(0, 2π3)\left( {0,\,\frac{{2\pi }}{3}} \right)
The correct answer is B ((0,,2π3)\left( {0,\\,\frac{{2\pi }}{3}} \right)) because the normal to the curve at x=π6x = \frac{\pi }{6} passes through this point.
449 Medium
449 of 932
Application +4 | -1
If y=sin⁡x⋅sin⁡2x⋅sin⁡3x…⋅sin⁡nxy=\sin x \cdot \sin 2 x \cdot \sin 3 x \ldots \cdot \sin n x, then dydx=……\frac{d y}{d x}=\ldots \ldots
View answer and explanation
Correct option: B
y∑k=1nkcot⁡kxy \sum\limits_{{k=1}}^n k \cot k x
Option B (y∑k=1nkcot⁡kxy \sum\limits_ {{k=1}}^n k \cot k x) is correct because by taking the logarithm of both sides and differentiating, we can express dy/dx in terms of y and a sum of cotangents.
450 Easy
450 of 932
Application +4 | -1
Let ff be differentiable for all xx. If f(1)=−2f\left( 1 \right) = - 2 and f′(x)≥2f'\left( x \right) \ge 2 for x∈[1,6]x \in \left[ {1,6} \right], then
View answer and explanation
Correct option: D
f(6)≥8f\left( 6 \right) \ge 8
Option D (f(6)≥8f\left( 6 \right) \ge 8) is correct because it is derived from the Mean Value Theorem, using the given condition on the derivative and the value of the function at x=1.
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