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JEE • MATHEMATICS • DIFFERENTIATION

Differentiation practice · Mixed difficulty

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301 Easy
301 of 932
Analyzing +4 | -1
The derivative of tan⁡−1(1+x2−1x)\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right) with respect to tan⁡−1(2x1−x21−2x2)\tan ^{-1}\left(\frac{2 x \sqrt{1-x^2}}{1-2 x^2}\right) at x=12x=\frac{1}{2}, is
View answer and explanation
Correct option: A
310\frac{\sqrt{3}}{10}
Option A (310\frac{\sqrt{3}}{10}) is correct because by substituting x=tan(θ)x = tan(\theta) and x=sin(ϕ)x = sin(\phi), we can simplify both inverse trigonometric functions and then find the derivative at x=1/2.
302 Easy
302 of 932
Application +4 | -1
The derivative of log⁡10x\log _{10} x with respect to x2x^2, is
View answer and explanation
Correct option: B
2x2log⁡10e\frac{2}{x^2} \log _{10} e
Option C is correct because it is derived by using the change of base formula and then differentiating with respect to x2.
303 Medium
303 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 (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)) is correct because it is the result of applying the chain rule to differentiate the composite function.
304 Easy
304 of 932
Application +4 | -1
The length of the subnormal of the curve y2=8ax(a>0)y^2=8 a x(a > 0) is -
View answer and explanation
Correct option: B
4a4 \mathrm{a}
Option B (4a4a) is correct because the length of the subnormal is calculated using the derivative of the curve's equation.
305 Medium
305 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 (- (d2y/dx2) * (dy/dx)-3) is correct because it is derived from the chain rule and inverse function differentiation.
306 Medium
306 of 932
Application +4 | -1
Let a function y=y(x)y=y(x) be defined parametrically by x=2t−∣t∣,y=t2+t∣t∣x=2 t-|t|, y=t^2+t|t|. Then y′(x),x>0y^{\prime}(x), x>0
View answer and explanation
Correct option: B
4x4 x
Option C (4x4x) is correct because after expressing yy in terms of xx for x>0x>0, differentiating yields y′(x)=4xy'(x) = 4x.
307 Easy
307 of 932
Analyzing +4 | -1
If f(x)=∣x−1∣f(x)=|x-1| and g(x)=f(f(f(x)))g(x)=f(f(f(x))), then for x>2,g′(x)x>2, g^{\prime}(x) is equal to
View answer and explanation
Correct option: C
−1-1 for 2<x<32 < x < 3
Option C (−1-1 for 2<x<32 < x < 3) is correct because for x>2x > 2, g(x)=3−xg(x) = 3 - x, and thus g′(x)=−1g'(x) = -1.
308 Medium
308 of 932
Application +4 | -1
If y=log⁡ax+log⁡xa+log⁡xx+log⁡aay=\log _a x+\log _x a+\log _x x+\log _a a, then dydx\frac{d y}{d x} is equal to
View answer and explanation
Correct option: D
1xlog⁡a−log⁡ax(log⁡x)2\frac{1}{x \log a}-\frac{\log a}{x(\log x)^2}
Option D (1xlog⁡a−log⁡ax(log⁡x)2\frac{1}{x \log a}-\frac{\log a}{x(\log x)^2}) is correct because after applying the change of base formula for logarithms and differentiating, the derivative is found to be 1xlog⁡a−log⁡ax(log⁡x)2\frac{1}{x \log a}-\frac{\log a}{x(\log x)^2}.
309 Easy
309 of 932
Application +4 | -1
If y=e12log⁡(1+tan⁡2x)y = {e^{\frac{1}{2}\log (1 + {{\tan }^2}x)}}, then dydx\frac{{dy}}{{dx}} is equal to
View answer and explanation
Correct option: C
sec⁡xtan⁡x\sec x\tan x
Option C (sec⁡xtan⁡x\sec x\tan x) is correct because, after simplifying the expression for y, the derivative dydx\frac{{dy}}{{dx}} is found to be sec⁡xtan⁡x\sec x\tan x.
310 Easy
310 of 932
Application +4 | -1
If f(x)=sin⁡(π3[x]−x2)f(x)=\sin \left(\frac{\pi}{3}[x]-x^2\right), where 2<x<32 < x < 3 and [.] represetns greatest integer function, then f′(5π3)f^{\prime}\left(\sqrt{\frac{5 \pi}{3}}\right) is equal to
View answer and explanation
Correct option: B
25π32 \sqrt{\frac{5 \pi}{3}}
Option B (25π32 \sqrt{\frac{5 \pi}{3}}) is correct because by substituting the greatest integer function and differentiating f(x)f(x), we can find f′(5π3)f'(\sqrt{\frac{5 \pi}{3}}).
311 Medium
311 of 932
Application +4 | -1
If function f(x)f(x) is differentiable at x=ax=a, then lim⁡x→ax2f(a)−a2f(x)x−a\lim _{x \rightarrow a} \frac{x^{2} f(a)-a^{2} f(x)}{x-a} is :
View answer and explanation
Correct option: C
2af(a)−a2f′(a)2 a f(a)-a^{2} f^{\prime}(a)
Option C (2af(a)−a2f′(a)2 a f(a)-a^{2} f^{\prime}(a)) is correct because it is the result of applying L'Hopital's rule to the given limit.
312 Medium
312 of 932
Application +4 | -1
If y=tan⁡−1(acos⁡x−bsin⁡xbcos⁡x+asin⁡x)dydx=y=\tan ^{-1}\left(\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right) \frac{d y}{d x}=
View answer and explanation
Correct option: C
-1
Option C (-1) is correct because after simplification using trigonometric identities and substitution, the derivative of y with respect to x is -1.
313 Medium
313 of 932
Application +4 | -1
If y=(1+x)(1+x2)(1+x4)…(1+x2m)y=(1+x)\left(1+x^2\right)\left(1+x^4\right) \ldots\left(1+x^{2 m}\right), then the value of dydx\frac{d y}{d x} at x=0x=0 is ......
View answer and explanation
Correct option: C
1
Option C is correct because the derivative evaluated at x=0 is 1.
314 Medium
314 of 932
Application +4 | -1
If y=log⁡x2+4(7x2−5x+1)y=\log _{x^2+4}\left(7 x^2-5 x+1\right), then dydx=\frac{d y}{d x}=
View answer and explanation
Correct option: A
1log⁡e(x2+4)(14x−57x2−5x+1−2xyx2+4)\frac{1}{\log _e\left(x^2+4\right)}\left(\frac{14 x-5}{7 x^2-5 x+1}-\frac{2 x y}{x^2+4}\right)
Option A is correct because it is the result of applying the change of base formula and the quotient rule for differentiation.
315 Medium
315 of 932
Analyzing +4 | -1
If f(x)f(x) is continuous and differentiable function such that f(1n)=0f\left(\frac{1}{n}\right)=0 for all n∈Nn \in N, then
View answer and explanation
Correct option: B
f(0)=0,f′(0)=0f(0)=0, f^{\prime}(0)=0
Option B (f(0)=0,f′(0)=0f(0)=0, f^{\prime}(0)=0) is correct because the given condition f(1n)=0f(\frac{1}{n})=0 for all n∈Nn \in N implies that f(0)=0f(0)=0 and f′(0)=0f^{\prime}(0)=0 due to continuity and differentiability.

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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
316 Easy
316 of 932
Application +4 | -1
If f(x)=∣x−2∣+∣x+1∣−x,f\left( x \right) = \left| {x - 2} \right| + \left| {x + 1} \right| - x, then f′(−10)f'\left( { - 10} \right) is equal to
View answer and explanation
Correct option: A
−3- 3
Option A (-3) is correct because for x < -1, the function simplifies to f(x) = -3x + 1, and thus f'(x) = -3.
317 Medium
317 of 932
Application +4 | -1
If x=cos⁡−1(11+t2),y=sin⁡−1(t1+t2)⇒dydxx = {\cos ^{ - 1}}\left( {\frac{1}{{\sqrt {1 + {t^2}} }}} \right),y = {\sin ^{ - 1}}\left( {\frac{t}{{\sqrt {1 + {t^2}} }}} \right) \Rightarrow \frac{{dy}}{{dx}} is equal to
View answer and explanation
Correct option: C
1
Option C (1) is correct because both x and y simplify to tan−1(t)tan^{-1}(t), making dydx\frac{{dy}}{{dx}} equal to 1.
318 Medium
318 of 932
Application +4 | -1
d2xdy2\frac{d^{2} x}{d y^{2}} equals :
View answer and explanation
Correct option: C
−(d2ydx2)(dydx)−3-\left(\frac{d^{2} y}{d x^{2}}\right)\left(\frac{d y}{d x}\right)^{-3}
Option C (-(d2ydx2\frac{{d^{2}y}}{{dx^{2}}})(dydx\frac{{dy}}{{dx}})−3^{-3}) is correct because it represents the correct expression for d2xdy2\frac{{d^{2}x}}{{dy^{2}}} in terms of derivatives of y with respect to x.
319 Easy
319 of 932
Application +4 | -1
Derivative of sin⁡x\sin x w.r.t. cos⁡x\cos x is
View answer and explanation
Correct option: C
−cot⁡x- \cot x
Option C (−cot⁡x- \cot x) is correct because it is the derivative of sin⁡x\sin x with respect to cos⁡x\cos x obtained using the chain rule.
320 Easy
320 of 932
Application +4 | -1
If y=sin⁡(msin⁡−1x)y = \sin \left( {m{{\sin }^{ - 1}}x} \right), then (1−x2)d2ydx2=\left( {1 - {x^2}} \right)\frac{{{d^2}y}}{{d{x^2}}} =
View answer and explanation
Correct option: B
xdydx−m2yx \frac{d y}{d x}-m^2 y
Option B (xdydx−m2yx \frac{d y}{d x}-m^2 y) is correct because it is the result of differentiating the given equation twice and rearranging the terms.
321 Easy
321 of 932
Application +4 | -1
The differential coefficient of f(sin⁡x)f(\sin x) with respect to xx, where f(x)=log⁡xf\left( x \right) = \log x, is
View answer and explanation
Correct option: B
cot⁡x\cot x
Option B (cot⁡x\cot x) is correct because the derivative of f(sin⁡x)f(\sin x) with respect to xx, where f(x)=log⁡xf(x) = \log x, is indeed cot⁡x\cot x.
322 Medium
322 of 932
Application +4 | -1
If y=aemx+be−axy=a e^{m x}+b e^{-a x}, then y2y_2 is equal to
View answer and explanation
Correct option: B
m2ym^2 y
Option B (m2ym^2 y) is correct because differentiating the given equation twice leads to this result.
323 Easy
323 of 932
Analyzing +4 | -1
The function f(x)f(x) given by f(x)= arc  sin(2x1+x2)f(x)=\,arc\,\,sin \left(\frac{2 x}{1+x^2}\right), is
View answer and explanation
Correct option: B
such that f′(x)={21+x2,    −1<x<1−21+x2,    ∣x∣>1f'(x) = \left\{ {\begin{array}{llllllllllllllllllll} {\frac{2}{{1 + {x^2}}},\,\,\,\, - 1 < x < 1}\\ {\frac{{ - 2}}{{1 + {x^2}}},\,\,\,\,|x| > 1} \end{array}} \right.
Option B is correct because the derivative of the arcsin function needs to be evaluated piecewise due to the range restriction of the arcsin function, resulting in different derivatives for |x| < 1 and |x| > 1.
324 Easy
324 of 932
Application +4 | -1
If g\mathrm{g} is the inverse of f\mathrm{f} and f′(x)=11+x3\mathrm{f}^{\prime}(x)=\frac{1}{1+x^3}, then find g′(x)g^{\prime}(x).
View answer and explanation
Correct option: C
1+[g(x)]31+[g(x)]^3
Option C (1+[g(x)]31+[g(x)]^3) is correct because by differentiating f[g(x)]=xf[g(x)] = x with respect to x and using the given f′(x)f'(x), we find that g′(x)=1+[g(x)]3g'(x) = 1 + [g(x)]^3.
325 Medium
325 of 932
Analyzing +4 | -1
The function f(x)=(x2−1)∣x2−3x+2∣+cos⁡(∣x∣)f(x)=\left(x^2-1\right)\left|x^2-3 x+2\right|+\cos (|x|) is not differentiable at
View answer and explanation
Correct option: D
2
Option D (2) is correct because the function is not differentiable at x=2 due to the absolute value function changing its behavior at that point.
326 Hard
326 of 932
Analyzing +4 | -1
If y2=p(x)y^2=p(x) is a polynomial of degree 3 , then 2ddx(y3⋅d2ydx2)2 \frac{d}{d x}\left(y^3 \cdot \frac{d^2 y}{d x^2}\right) is equal to
View answer and explanation
Correct option: C
p(x)⋅p′′′(x)p(x) \cdot p^{\prime \prime \prime}(x)
Option C (p(x) \cdot p^{\'\''}(x)p(x) \cdot p^{\'\''}(x)) is correct because it is the result of differentiating the given expression and simplifying using the given condition.
327 Medium
327 of 932
Application +4 | -1
d2xdy2\frac{d^{2} x}{d y^{2}} equals
View answer and explanation
Correct option: C
−(d2ydx2)(dydx)−3-\left(\frac{d^{2} y}{d x^{2}}\right)\left(\frac{d y}{d x}\right)^{-3}
The correct answer is C (−(d2ydx2)(dydx)−3-\left(\frac{d^{2} y}{d x^{2}}\right)\left(\frac{d y}{d x}\right)^{-3}) because it represents the second derivative of x with respect to y in terms of derivatives of y with respect to x.
328 Easy
328 of 932
Application +4 | -1
If y=2x.32x−1,y = {2^x}{.3^{2x - 1}}, then d2ydx2\frac{{{d^2}y}}{{d{x^2}}} is equal to
View answer and explanation
Correct option: D
(log⁡18)2y{(\log 18)^2}y
Option D ((log⁡18)2y{(\log 18)^2}y) is correct because the second derivative of the given function can be expressed in terms of the original function and the square of the logarithm of 18.
329 Medium
329 of 932
Application +4 | -1
If x=a(θ−sin⁡θ),y=a(1−cos⁡θ),x = a(\theta - \sin \theta ),y = a\left( {1 - \cos \theta } \right), then dydx\frac{{dy}}{{dx}} is equal to
View answer and explanation
Correct option: A
cot⁡θ2\cot \frac{\mathrm{\theta }}{2}
Option A (cot⁡θ2\cot \frac{\mathrm{\theta }}{2}) is the correct answer because it is the simplified form of dydx\frac{{dy}}{{dx}} after differentiating xx and yy with respect to θ\theta and applying trigonometric identities.
330 Easy
330 of 932
Application +4 | -1
The differential coefficient of f(log⁡x),f\left( {\log x} \right), where  f(x)=log⁡x\,f\left( x \right) = \log x is
View answer and explanation
Correct option: B
(xlog⁡x)−1{(x\log x)^{ - 1}}
Option B ((xlog⁡x)−1{(x\log x)^{ - 1}}) is correct because the derivative of log⁡(log⁡x)\log(\log x) is 1xlog⁡x\frac{1}{{x\log x}}.
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