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JEE • MATHEMATICS • DIFFERENTIATION

Differentiation practice · Mixed difficulty

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Showing 121-150 of 932 questions

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121 Easy
121 of 932
Application +4 | -1
Let f:R→Rf: R \rightarrow R be a function satisfying f(x+y)=f(x)+2y2+kxyf(x+y)=f(x)+2 y^2+k x y for all x,y∈Rx, y \in R. If f(1)=2f(1)=2 and f(2)=8f(2)=8, then f(x)f(x) is equal to
View answer and explanation
Correct option: A
2x22 x^2
Option A is correct because using the given condition f(x+y)=f(x)+2y2+kxyf(x+y)=f(x)+2 y^2+k x y and the values f(1)=2f(1)=2 and f(2)=8f(2)=8, we find f(x)=2x2f(x)=2 x^2.
122 Easy
122 of 932
Application +4 | -1
ddx(cos⁡−11+cos⁡x2)\frac{d}{d x}\left(\cos ^{-1} \sqrt{\frac{1+\cos x}{2}}\right) is equal to
View answer and explanation
Correct option: B
12\frac{1}{2}
Option B (12\frac{1}{2}) is correct because the expression simplifies to x/2x/2 and its derivative is 1/21/2.
123 Medium
123 of 932
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.
124 Medium
124 of 932
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.
125 Medium
125 of 932
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.
126 Easy
126 of 932
Application +4 | -1
If y=sin⁡−1(5x+121−x213)y=\sin ^{-1}\left(\frac{5 x+12 \sqrt{1-x^2}}{13}\right), then dydx\frac{d y}{d x} is equal to:
View answer and explanation
Correct option: B
11−x2\frac{1}{\sqrt{1-x^2}}
Option B (11−x2\frac{{1}}{{\sqrt{{1-x^2}}}}) is correct because it is the derivative of the given function after simplification using trigonometric substitution.
127 Easy
127 of 932
Analyzing +4 | -1
If ddx(f(x))=g(x);ddx(g(x))=f(x2)\frac{d}{d x}(f(x))=g(x) ; \frac{d}{d x}(g(x))=f\left(x^{2}\right), then d2dx2f(x3)\frac{d^{2}}{d x^{2}} f\left(x^{3}\right) can be expressed in the form kxaf(xb)+pxg(xx)k x^{a} f\left(x^{b}\right)+p x g\left(x^{x}\right) where a,b,c,k,p∈Na, b, c, k, p \in N, then the value of (k+p+a+b+c)(k+p+a+b+c) equal to
View answer and explanation
Correct option: C
28
Option C (28) is correct because after finding the second derivative of f(x3) and comparing it to the given form, the sum of the coefficients and exponents is 28.
128 Medium
128 of 932
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.
129 Hard
129 of 932
Analyzing +4 | -1
If y=cos⁡−1cos⁡(∣x∣−f(x))y = {\cos ^{ - 1}}\cos \left( {\left| x \right| - f\left( x \right)} \right), where
f(x)={1,if  x>0−1,if  x<00,if  x=0f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {1,if\,\,x > 0}\\ { - 1,if\,\,x < 0}\\ {0,if\,\,x = 0} \end{array}} \right. Then, (dy/dx)x=5π4\left( {dy/dx} \right)x = \frac{{5\pi }}{4} is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because, for x > 0, y = cos⁻¹(cos(x-1)), and for x in the interval [1, π+1], y = x-1, so dy/dx = 1.
130 Easy
130 of 932
Application +4 | -1
Let f(x)=x3+4x2+6xf(x)=x^3+4 x^2+6 x and g(x)g(x) be its inverse then the value of g′(−4)g^{\prime}(-4) :
View answer and explanation
Correct option: C
12\frac{1}{2}
Option C (12\frac{1}{2}) is correct because it is found by using the inverse function derivative formula and evaluating at the correct point.
131 Easy
131 of 932
Analyzing +4 | -1
The derivative of cos⁡−1(1−x21+x2){\cos ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right) with respect to cot⁡−1(1−3x23x−x3){\cot ^{ - 1}}\left( {\frac{{1 - 3{x^2}}}{{3x - {x^3}}}} \right) is
View answer and explanation
Correct option: C
23\frac{2}{3}
Option C (23\frac{2}{3}) is correct because by using the substitution x=tan⁡θx = \tan \theta, we can simplify the expressions and find that the derivative of y1{y_1} with respect to y2{y_2} is 23\frac{2}{3}.
132 Medium
132 of 932
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.
133 Easy
133 of 932
Application +4 | -1
A spherical balloon is expanding. If the radius is increasing at the rate of 5 inch per minute, the rate at which the volume increases (in cubic inches per minute) when the radius is 10 inch is
View answer and explanation
Correct option: C
2000π2000 \pi
Option C (2000π2000 \pi) is correct because it represents the rate of change of the volume of the sphere when the radius is 10 inches and increasing at 5 inches per minute.
134 Easy
134 of 932
Analyzing +4 | -1
The range of a∈Ra \in R for which the function
f(x)=(4a−3)(x+log⁡e5)+2(a−7)cot⁡(x2)sin⁡2(x2),  x≠2nπ,  n∈Nf\left( x \right) = \left( {4a - 3} \right)\left( {x + {{\log }_e}5} \right) + 2\left( {a - 7} \right)\cot \left( {\frac{x}{2}} \right){\sin ^2}\left( {\frac{x}{2}} \right),\,\,x \ne 2n\pi ,\,\,n \in N has critical points, is:
View answer and explanation
Correct option: A
[−43, 2]\left[ { - \frac{4}{3},\,2} \right]
Option A ([−43, 2]\left[ { - \frac{4}{3},\,2} \right]) is correct because it represents the range of 'a' for which the given function has critical points, derived from the condition -1 <= cos(x) <= 1.
135 Easy
135 of 932
Application +4 | -1
If y=sin⁡x+sin⁡x+sin⁡x+...∞,y = \sqrt {\sin x + \sqrt {\sin x + \sqrt {\sin x + ...\infty ,} } }, then dydx\frac{{dy}}{{dx}} is equal to
View answer and explanation
Correct option: A
cos⁡x2y−1\frac{{\cos x}}{{2y - 1}}
Option A (cos x / (2y - 1)) is correct because squaring the equation and then differentiating implicitly gives cos x / (2y - 1).

Explore Other Chapters in MATHEMATICS

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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
136 Easy
136 of 932
Application +4 | -1
If y=cos⁡x2+cos⁡x2+cos⁡x2+⋯+∞y=\sqrt{\cos x^2+\sqrt{\cos x^2+\sqrt{\cos x^2+\cdots+\infty}}}, then dydx\frac{d y}{d x}
View answer and explanation
Correct option: B
−2xsin⁡x22y−1\frac{-2 x \sin x^2}{2 y-1}
Option B (−2xsin⁡x22y−1\frac{-2 x \sin x^2}{2 y-1}) is correct because differentiating the given equation implicitly leads to this result.
137 Easy
137 of 932
Application +4 | -1
d2xdy2\frac{d^2 x}{d y^2} equals
View answer and explanation
Correct option: A
−(d2ydx2)−1(dydx)−3-\left(\frac{d^2 y}{d x^2}\right)^{-1}\left(\frac{d y}{d x}\right)^{-3}
Option A (−(d2ydx2)−1(dydx)−3-\left(\frac{{d^2 y}}{{d x^2}}\right)^{-1}\left(\frac{{d y}}{{d x}}\right)^{-3}) is correct because it represents the second derivative of x with respect to y in terms of derivatives of y with respect to x.
138 Easy
138 of 932
Application +4 | -1
If y=log⁡{(1+x1−x)14}−12tan⁡−1xy = \log \left\{ {{{\left( {\frac{{1 + x}}{{1 - x}}} \right)}^{\frac{1}{4}}}} \right\} - \frac{1}{2}{\tan ^{ - 1}}x, then dydx=\frac{{dy}}{{dx}} =
View answer and explanation
Correct option: B
x21−x4\frac{{{x^2}}}{{1 - {x^4}}}
Option B (x21−x4\frac{{{x^2}}}{{1 - {x^4}}}) is correct because it is the simplified derivative after applying logarithmic and inverse trigonometric differentiation rules.
139 Medium
139 of 932
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}}}}.
140 Medium
140 of 932
Application +4 | -1
If f′(x)=g(x)f'(x) = g(x) and g′(x)=−f(x)g'(x) = - f(x) for all real xx and f(5)=2=f′(5),f(5) = 2 = f'(5), then f2(10)+g2(10){f^2}(10) + {g^2}(10) equals
View answer and explanation
Correct option: C
8
Option C (8) is correct because the expression f2(10)+g2(10){f^2}(10) + {g^2}(10) is constant and equal to f2(5)+g2(5){f^2}(5) + {g^2}(5), which evaluates to 8.
141 Easy
141 of 932
Application +4 | -1
If yy is a function of xx and log⁡(x+y)=2xy\log (x+y)=2 x y, then (dydx)x=0\left(\frac{d y}{d x}\right)_{x=0} is
View answer and explanation
Correct option: A
1
Option A (1) is correct because differentiating the given equation and substituting x=0 yields dy/dx = 1.
142 Easy
142 of 932
Application +4 | -1
If f′(x)=sin⁡(log⁡x) f'\left( x \right) = \sin (\log x)\, and y=f(2x+33−2x)y = f\left( {\frac{{2x + 3}}{{3 - 2x}}} \right), then dydx\frac{{dy}}{{dx}} at x=1x = 1 is equal to
View answer and explanation
Correct option: C
12sinlog(5)12\mathrm{ }sin\mathrm{ }log\mathrm{ }\left( 5 \right)
Option C (12sinlog(5)12\mathrm{ }sin\mathrm{ }log\mathrm{ }\left( 5 \right)) is correct because it is the value of the derivative of the composite function evaluated at x=1x = 1.
143 Easy
143 of 932
Application +4 | -1
Let solution of differential equation e−x(dy−ℓn(ex)dx)−dx=0\mathrm{e}^{-\mathrm{x}}(\mathrm{dy}-\ell \mathrm{n}(\mathrm{ex}) \mathrm{dx})-\mathrm{dx}=0 is given by y=f(x)\mathrm{y}=\mathrm{f}(\mathrm{x}). If f(1)=e+1\mathrm{f}(1)=\mathrm{e}+1, then value of lim⁡x→0+f(x)\lim \limits_{\mathrm{x} \rightarrow 0^{+}} \mathrm{f}(\mathrm{x}) is
View answer and explanation
Correct option: A
2
Option A (2) is correct. The solution of the differential equation is found, the initial condition is applied to find the constant, and then the limit is evaluated.
144 Medium
144 of 932
Application +4 | -1
Let f(x)=ex,g(x)=e−xf(x)=e^x, g(x)=e^{-x} and h(x)=g(f(x))h(x)=g(f(x)). Then h′(0)h^{\prime}(0) is equal to
View answer and explanation
Correct option: A
−1e\frac{-1}{e}
Option A (-1/e) is correct because it is the value of the derivative of h(x) = e(-e^x) evaluated at x=0.
145 Medium
145 of 932
Analyzing +4 | -1
If y=f(x)y=f(x) and x=g(y)x=g(y) are inverse functions of each other, then
View answer and explanation
Correct option: B
g′′(y)=−f′′(x)(f′(x))3g^{\prime \prime}(y)=-\frac{f^{\prime \prime}(x)}{\left(f^{\prime}(x)\right)^3}
Option B (g′′(y)=−f′′(x)(f′(x))3g^{\prime \prime}(y)=-\frac{f^{\prime \prime}(x)}{\left(f^{\prime}(x)\right)^3}) is correct because it is the correct second derivative of the inverse function, derived using the chain rule.
146 Easy
146 of 932
Analyzing +4 | -1
x=cosθ,y=sin⁡5θ⇒(1−x2)d2ydx2−xdydxx = \mathrm{cos\theta },y = \sin 5\mathrm{\theta } \Rightarrow \left( {1 - {x^2}} \right)\frac{{{d^2}y}}{{d{x^2}}} - x\frac{{dy}}{{dx}} is
View answer and explanation
Correct option: D
−25y- 25y
Option D (-25y) is correct because after finding the first and second derivatives and substituting, the expression simplifies to -25y.
147 Medium
147 of 932
Application +4 | -1
If sin⁡(x+y)=log⁡(x+y)\sin (x+y)=\log (x+y), then dydx\frac{d y}{d x} is equal to
View answer and explanation
Correct option: D
-1
Option D is correct because the derivative of the implicit function simplifies to -1.
148 Medium
148 of 932
Application +4 | -1
If y=cos⁡−1(1−log⁡x1+log⁡x),y = {\cos ^{ - 1}}\left( {\frac{{1 - \log x}}{{1 + \log x}}} \right), then dydx\frac{{dy}}{{dx}} at x=ex = e
View answer and explanation
Correct option: C
12e\frac{1}{{2e}}
Option C (12e\frac{1}{{2e}}) is correct because after differentiating and substituting x=ex = e, the derivative evaluates to 12e\frac{1}{{2e}}.
149 Medium
149 of 932
Application +4 | -1
Let f(x+y2)=f(x)+f(y)2f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2} for all real xx and yy for f′(0)=−1,f(0)=1f^{\prime}(0)=-1, f(0)=1, then f′(2)=f^{\prime}(2)=
View answer and explanation
Correct option: C
-1
Option C (-1) is correct because the given functional equation implies that f′(x)f'(x) is constant, and since f′(0)=−1f'(0) = -1, then f′(2)=−1f'(2) = -1.
150 Medium
150 of 932
Application +4 | -1
Differential co-efficient of sin⁡−1x\sin ^{-1} x w.r.t cos⁡−11−x2\cos ^{-1} \sqrt{1-x^2} is
View answer and explanation
Correct option: A
1
Option A (1) is correct because the derivative of sin⁡−1x\sin^{-1}x with respect to cos⁡−11−x2\cos^{-1}\sqrt{1-x^2} is 1, as they are essentially the same function.
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