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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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331 Medium
331 of 932
Application +4 | -1
Let y=y(x)y=y(x) be aa function of xx satisfying y1−x2=k−x1−y2y \sqrt{1-x^{2}}=k-x \sqrt{1-y^{2}} where kk is aa constant and y(12)=−14y\left(\frac{1}{2}\right)=-\frac{1}{4}. Then dydx\frac{d y}{d x} at x=12x=\frac{1}{2}, is equal to:
View answer and explanation
Correct option: B
−52-\frac{\sqrt{5}}{2}
Option B (−52-\frac{\sqrt{5}}{2}) is correct because it is the value of dydx\frac{dy}{dx} obtained by implicit differentiation and substitution of the given values.
332 Easy
332 of 932
Application +4 | -1
If x=f(t),y=g(t)x=f(t), y=g(t), then d2ydx2=……\frac{d^2 y}{d x^2}=\ldots \ldots
View answer and explanation
Correct option: B
f′g′′−g′f′′(f)3\frac{{{f^\prime }g'' - {g^\prime }{f^{\prime \prime }}}}{{{{(f)}^3}}}
Option B is correct because it represents the correct formula for the second derivative of y with respect to x when both x and y are functions of t.
333 Easy
333 of 932
Application +4 | -1
f(x)f\left( x \right) is a polynomial of degree 2, f(0)=4,f′(0)=3f\left( 0 \right) = 4,f'\left( 0 \right) = 3 and f′′(0)=4f''\left( 0 \right) = 4, then f(−1)f\left( { - 1} \right) is equal to
View answer and explanation
Correct option: A
3
Option A (3) is correct because given the conditions, the polynomial is determined to be f(x)=2x2+3x+4f(x) = 2x^2 + 3x + 4, and f(−1)=3f(-1) = 3.
334 Easy
334 of 932
Application +4 | -1
If 2x2−3xy+y2+x+2y−8=0,2{x^2} - 3xy + {y^2} + x + 2y - 8 = 0, then dydx\frac{{dy}}{{dx}} is equal to
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.
335 Easy
335 of 932
Application +4 | -1
Let f:R→Rf: R \rightarrow R and g:R→Rg: R \rightarrow R be differentiable functions such that f(x)=x3+3x+2,g(f(x))=xf(x)=x^3+3 x+2, g(f(x))=x for all x∈Rx \in R. Then, g′(2)=g^{\prime}(2)=
View answer and explanation
Correct option: C
13\frac{1}{3}
Option C (13\frac{{1}}{{3}}) is correct because it is derived by differentiating g(f(x))=xg(f(x)) = x and substituting x=0x=0.
336 Medium
336 of 932
Analyzing +4 | -1
If f(x)=cos⁡xcos⁡2xcos⁡22xcos⁡23x……f(x)=\cos x \cos 2 x \cos 2^2 x \cos 2^3 x \ldots \ldots cos⁡2n−1x\cos 2^{n-1} x and n>1n>1, then f′(π2)f^{\prime}\left(\frac{\pi}{2}\right), is
View answer and explanation
Correct option: A
1
Option D (None of these) is correct because f′(x)=0f'(x) = 0 for x<0x < 0, f′(−2)=0f'(-2) = 0 and f′(−1)f'(-1) is not defined.
337 Medium
337 of 932
Application +4 | -1
The derivative of y=xln⁡xy=x^{\ln x} is
View answer and explanation
Correct option: C
2xln⁡x−1ln⁡x2 x^{\ln x-1} \ln x
Option C is correct because it is the result of logarithmic differentiation.
338 Easy
338 of 932
Application +4 | -1
On the curve y=x3y=x^3, the point at which the tangent line in parallel to the chord through the point (−1,−1)(-1,-1) and (2,8)(2,8) is
View answer and explanation
Correct option: A
(1,1)(1,1)
Option A ((1,1)) is correct because the slope of the tangent at that point is equal to the slope of the chord connecting (-1,-1) and (2,8).
339 Easy
339 of 932
Application +4 | -1
If f(x)=(1+x)nf(x)=(1+x)^n, then the value of f(0)+f(0)+ f′(0)+12!f′′(0)+⋯+1n!fn(0)f^{\prime}(0)+\frac{1}{2 !} f^{\prime \prime}(0)+\cdots+\frac{1}{n !} f^n(0) is
View answer and explanation
Correct option: B
2n2 n
Option D (none of these) is correct because the expression simplifies to 2n2^n, not 2n2n.
340 Easy
340 of 932
Analyzing +4 | -1
If f(x)=xnf(x)=x^{n}, then the value of
f(1)−f′(1)1!+f′′(1)2!−f′′′(1)3!+………..(−1)nfn(1)n!f(1)-\frac{f^{\prime}(1)}{1 !}+\frac{f^{\prime \prime}(1)}{2 !}-\frac{f^{\prime \prime \prime}(1)}{3 !}+\ldots \ldots \ldots . . \frac{(-1)^{n} f^{n}(1)}{n !} is
View answer and explanation
Correct option: D
0 .
Option D (0) is correct because the given expression is the binomial expansion of (1-1)^n which equals 0.
341 Medium
341 of 932
Analyzing +4 | -1
Let f,g:R→R\mathrm{f}, \mathrm{g}: \mathrm{R} \rightarrow \mathrm{R} be two functions defined by
f(x)={xsin⁡(1x),x≠00,,x=0, and g(x)=xf(x)f(x)=\left\{\begin{array}{ll} x \sin \left(\frac{1}{x}\right), x \neq 0 \\ 0, & , x=0 \end{array}, \text { and } g(x)=x f(x)\right.

Statement I: ff is a continuous function at x=0\mathrm{x}=0.
Statement II: gg is a differentiable function at x=0\mathrm{x}=0.
View answer and explanation
Correct option: C
Statement I\mathrm{I} is true, statement II is false.
Option C is correct because f(x) is continuous at x=0, but g(x) is not differentiable at x=0.
342 Medium
342 of 932
Analyzing +4 | -1
Let f(θ)=sin⁡(tan⁡−1(sin⁡θcos⁡2θ))f(\theta)=\sin \left(\tan ^{-1}\left(\frac{\sin \theta}{\sqrt{\cos 2 \theta}}\right)\right), where −π4<θ<π4-\frac{\pi}{4}<\theta<\frac{\pi}{4}. Then, the value of dd(tan⁡θ)⋅f(θ)\frac{d}{d(\tan \theta)} \cdot f(\theta), is
View answer and explanation
Correct option: A
1
Option A (1) is correct because after simplification, f(θ)=tan⁡θf(\theta) = \tan \theta, and the derivative of tan⁡θ\tan \theta with respect to tan⁡θ\tan \theta is 1.
343 Easy
343 of 932
Application +4 | -1
If cos⁡(x+y)=ysin⁡x\cos (x+y)=y \sin x, then dydx\frac{d y}{d x} is equal to
View answer and explanation
Correct option: A
−sin⁡(x+y)+ycos⁡xsin⁡x+sin⁡(x+y)-\frac{\sin (x+y)+y \cos x}{\sin x+\sin (x+y)}
Option A is correct because it is the result of implicit differentiation of the given equation and solving for dy/dx.
344 Easy
344 of 932
Application +4 | -1
ddx[e−ax2log⁡(sin⁡x)]\frac{d}{d x}\left[e^{-a x^{2}} \log (\sin x)\right]
View answer and explanation
Correct option: C
e−ax2(cot⁡x−2axlog⁡sin⁡x)e^{-a x^{2}}(\cot x-2 a x \log \sin x)
Option C is correct because the derivative is found using the product rule and simplifying the resulting expression.
345 Easy
345 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} + {\mathrm{y}^4} = {\mathrm{t}^2} + \frac{1}{{{\mathrm{t}^2}}}, then dydx\frac{{dy}}{{dx}} is equal to
View answer and explanation
Correct option: B
−yx- \frac{y}{x}
Option B ( −yx\ - \frac{y}{x}) is correct because by manipulating the given equations, we find that x2y2=1{x^2}{y^2} = 1, and differentiating this yields dydx=−yx\frac{{dy}}{{dx}} = - \frac{y}{x}.
346 Medium
346 of 932
Analyzing +4 | -1
If y=(log⁡cos⁡xsin⁡x)(log⁡sin⁡xcos⁡x)+sin⁡−12x1+x2,y = ({\log _{\cos x}}\sin x)({\log _{\sin x}}\cos x) + {\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}}, then dydx\frac{{\mathrm{d}y}}{{\mathrm{d}x}} at x=π2x = \frac{\pi }{2} is equal to
View answer and explanation
Correct option: D
None of the above
Option D (None of the above) is correct because the function is not defined at x=π2x = \frac{\pi}{2}, so the derivative cannot be determined.
347 Medium
347 of 932
Application +4 | -1
The function f(x)=ex+xf(x)=e^x+x, being differentiable and one-one, has a differentiable inverse f−1(x)f^{-1}(x). The value of ddx(f−1)\frac{d}{d x}\left(f^{-1}\right) at the point f(log⁡2)f(\log 2) is
View answer and explanation
Correct option: B
13\frac{1}{3}
Option B (13\frac{1}{3}) is correct because it is the value of the derivative of the inverse function at the specified point, calculated using the derivative of inverse functions.
348 Easy
348 of 932
Application +4 | -1
A particle is moving along the parabola y2=4(x+2)y^2=4(x+2). As it passes through the point (7,6)(7,6) its yy-coordinate is increasing at the rate of 3 units per second. The rate at which xx-coordinate change at this instant is (in units/sec)
View answer and explanation
Correct option: D
9
Option D (9) is correct because it's the rate of change of the x-coordinate, calculated using implicit differentiation.
349 Easy
349 of 932
Application +4 | -1
The derivative of f(x)=3∣2+x∣f(x)=3|2+x| at the point x0=−3x_{0}=-3 is
View answer and explanation
Correct option: B
-3
Option B (-3) is correct because the derivative of the function at x=-3 is calculated considering the absolute value function.
350 Easy
350 of 932
Application +4 | -1
Let ff be a bijection satisfying f′(x)=f(x)f^{\prime}(x)=f(x). Then, (f−1)′′(x)\left(f^{-1}\right)^{\prime \prime}(x) is equal to
View answer and explanation
Correct option: B
−1x2-\frac{1}{x^2}
Option B is correct because it correctly applies the derivative rules for inverse functions and uses the given condition f'(x) = f(x).
351 Easy
351 of 932
Application +4 | -1
Let f:R→Rf: R \rightarrow R be a function given by f(x+y)=f(x)f(y)f(x+y)=f(x) f(y) for all x,y∈Rx, y \in R

If f(x)≠0f(x) \neq 0 for all x∈Rx \in R and f′(0)f^{\prime}(0) exists, then f′(x)f^{\prime}(x) equals
View answer and explanation
Correct option: B
f(x)f′(0)f(x) f^{\prime}(0) for all x∈Rx \in R
Option B is correct because differentiating f(x+y)=f(x)f(y)f(x+y)=f(x)f(y) leads to f′(x)=f(x)f′(0)f'(x) = f(x)f'(0) for all x∈Rx \in R.
352 Easy
352 of 932
Application +4 | -1
If f(x+y)=f(x)⋅f(y)∀x⋅yf(x+y)=f(x) \cdot f(y) \forall x \cdot y and f(5)=2f(5)=2, f′(0)=3f^{\prime}(0)=3, then f′(5)f^{\prime}(5) is
View answer and explanation
Correct option: C
6
Option C (6) is correct because f′(5)f'(5) is calculated to be 6 using the given functional equation and derivative at 0.
353 Medium
353 of 932
Analyzing +4 | -1
If x2+y2+sin⁡y=4x^{2}+y^{2}+\sin y=4, then the value of d2ydx2\frac{d^{2} y}{d x^{2}} at the point (−2,0)(-2,0) is
View answer and explanation
Correct option: A
-34
Option A (-34) is correct because it is the value of the second derivative of y with respect to x, evaluated at the point (-2, 0).
354 Easy
354 of 932
Application +4 | -1
If f(x)=log⁡∣2x∣,x≠0f(x)=\log |2 x|, x \neq 0 then f′(x)f^{\prime}(x) is equal to
View answer and explanation
Correct option: A
11\frac{1}{1}
The correct answer is D (None of these). The derivative of f(x)=log⁡∣2x∣f(x) = \log |2x| is 1x\frac{{1}}{{x}} for all x≠0x \neq 0. However, the provided option A is also 1x\frac{{1}}{{x}}, so there is a typo in the options.
355 Easy
355 of 932
Application +4 | -1
ddxcos⁡−1x−x−1x+x−1\frac{d}{d x} \cos ^{-1} \frac{x-x^{-1}}{x+x^{-1}} is equal to
View answer and explanation
Correct option: D
−21+x2\frac{-2}{1+x^{2}}
Option D (−21+x2\frac{{-2}}{{1+x^{2}}}) is correct because by substituting x=cot⁡θx = \cot \theta, the expression simplifies to 2θ2\theta, and its derivative is −21+x2\frac{{-2}}{{1+x^{2}}}.
356 Easy
356 of 932
Analyzing +4 | -1
If f(x)=xn,f\left( x \right) = {x^n}, then the value of f(1)−f′(1)1!+f′′(1)2!−f′′′(1)3!+…+(−1)nfn(1)n!f\left( 1 \right) - \frac{{f'\left( 1 \right)}}{{1!}} + \frac{{f''\left( 1 \right)}}{{2!}} - \frac{{f'''\left( 1 \right)}}{{3!}} + \ldots + \frac{{{{\left( { - 1} \right)}^n}{f^n}\left( 1 \right)}}{{n!}} is
View answer and explanation
Correct option: C
0
Option C (0) is correct because the expression is equivalent to the binomial expansion of (1-1)^n which equals 0.
357 Easy
357 of 932
Analyzing +4 | -1
Let ff be a twice differentiable function such that f′′(x)=−f(x)f''\left( x \right) = - f\left( x \right) and f′(x)=g(x).f'\left( x \right) = \mathrm{g}\left( x \right). If h′(x)=[f(x)2+g(x)2]h(1)=8h'\left( x \right) = \left[ {f{{\left( x \right)}^2} + \mathrm{g}{{\left( x \right)}^2}} \right]h\left( 1 \right) = 8 and h(0)=2,h\left( 0 \right) = 2, then h(2)h\left( 2 \right) is equal to
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because after solving the differential equation and using the given conditions, we find that h(2)=14h(2) = 14, which is not among the provided options.
358 Easy
358 of 932
Understanding +4 | -1
Which of the following is/are true?

I. The derivative of sin⁡(x+a)\sin (x+a) is cos⁡(x+a)\cos (x+a), where aa is a fixed non-zero constant.

II. The derivative of cosec⁡xcot⁡x\operatorname{cosec} x \cot x is cosec⁡3x−cot⁡2xcosec⁡x\operatorname{cosec}^3 x-\cot ^2 x \operatorname{cosec} x
View answer and explanation
Correct option: B
Only I is true
Option B (Only I is true) is correct because the derivative of sin(x+a) is cos(x+a), while the derivative of cosec(x)cot(x) is -cosec3(x) - cot2(x)cosec(x).
359 Medium
359 of 932
Analyzing +4 | -1
If f′′(x)=−f(x)f''\left( x \right) = - f\left( x \right), where f(x)f\left( x \right) is a continuous double differentiable function and g(x)=f′(x). \mathrm{g}\left( x \right) = f'\left( x \right).\, If F(x)=(f(x2))2+(g(x2))2F\left( x \right) = {\left( {f\left( {\frac{x}{2}} \right)} \right)^2} + {\left( {\mathrm{g}\left( {\frac{x}{2}} \right)} \right)^2} and F(5)=5F\left( 5 \right) = 5 then F(10)F\left( {10} \right)
View answer and explanation
Correct option: B
5
Option B (5) is correct because F′(x)=0F'(x) = 0, indicating that F(x)F(x) is a constant function. Therefore, F(10)=F(5)=5F(10) = F(5) = 5.
360 Medium
360 of 932
Application +4 | -1
If u=log⁡(x2+y2x+y),u = \log \left( {\frac{{{x^2} + {y^2}}}{{x + y}}} \right), then the value of x∂u∂x+y∂u∂yx\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} is
View answer and explanation
Correct option: C
1
Option C (1) is correct because the expression x∂u∂x+y∂u∂yx\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} simplifies to 1 using Euler's theorem for homogeneous functions.
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