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JEE • MATHEMATICS • DIFFERENTIATION

Differentiation practice · Medium

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Showing 361-390 of 402 questions

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361 Medium
361 of 402
Analyzing +4 | -1
If y=tan⁡−1(log⁡(e/x2)log⁡(ex2))+tan⁡−1(3+2log⁡x1−6log⁡x)y = {\tan ^{ - 1}}\left( {\frac{{\log \left( {e/{x^2}} \right)}}{{\log \left( {e{x^2}} \right)}}} \right) + {\tan ^{ - 1}}\left( {\frac{{3 + 2\log x}}{{1 - 6\log x}}} \right), then d2ydx2\frac{{{d^2}y}}{{d{x^2}}} is equal to
View answer and explanation
Correct option: C
0
Option C (0) is correct because after simplification, y becomes a constant, and the second derivative of a constant is zero.
362 Medium
362 of 402
Application +4 | -1
If y=tan⁡−11+x2−1−x21+x2+1−x2,y = {\tan ^{ - 1}}\frac{{\sqrt {1 + {x^2}} - \sqrt {1 - {x^2}} }}{{\sqrt {1 + {x^2}} + \sqrt {1 - {x^2}} }}, then dydx\frac{{dy}}{{dx}} is equal to
View answer and explanation
Correct option: D
x1−x4\frac{x}{{\sqrt {1 - {x^4}} }}
Option D (x1−x4\frac{x}{{\sqrt {1 - {x^4}} }}) is correct because it is the result of simplifying the expression using trigonometric substitution and then differentiating.
363 Medium
363 of 402
Analyzing +4 | -1
If f(x)=∫0x(f(t))2dt,f:R→Rf(x)=\int_0^x(f(t))^2 d t, f: R \rightarrow R be differentiable function and f(g(x))f(g(x)) is differentiable function at x=ax=a, then
View answer and explanation
Correct option: A
g(x)g(x) must be differentiable at x=ax=a
Option A (0) is correct because, after substitution, the integral becomes the integral of an odd function over a symmetric interval, which is zero.
364 Medium
364 of 402
Application +4 | -1
If xmyn=(x+y)m+n{x^m}{y^n} = {\left( {x + y} \right)^{m + n}}, then (dy/dx)x=1,y=2{\left( {dy/dx} \right)_{x = 1,y = 2}} is equal to
View answer and explanation
Correct option: B
2
Option B (2) is correct because implicit differentiation and substituting x=1, y=2 leads to dy/dx = 2.
365 Medium
365 of 402
Application +4 | -1
If gg is inverse of ff and f′(x)=11+xnf^{\prime}(x)=\frac{1}{1+x^n}, then g′(x)g^{\prime}(x) equals
View answer and explanation
Correct option: C
1+[g(x)]n1+[g(x)]^n
Option C is correct because it correctly applies the inverse function derivative formula and substitutes the given function's derivative.
366 Medium
366 of 402
Application +4 | -1
If G(x)=125−x2\mathrm{G}(\mathrm{x})=\frac{1}{\sqrt{25-\mathrm{x}^2}}, then xx→0G(x+h)−G(4)x−4x_{x \rightarrow 0} \frac{\mathrm{G}(\mathrm{x}+\mathrm{h})-\mathrm{G}(4)}{\mathrm{x}-4} has the value
View answer and explanation
Correct option: D
none of these
Option D (none of these) is correct because the limit represents the derivative of G(x) at x=4, and the calculated derivative does not match any of the given options.
367 Medium
367 of 402
Application +4 | -1
The interval of increase of the function y=y= x−2sin⁡xx-2 \sin x if 0≤x≤2π0 \leq x \leq 2 \pi, is
View answer and explanation
Correct option: D
(π/3,5π/3)(\pi / 3,5 \pi / 3)
Option D (π3,5π3\frac{{\pi}}{{3}}, \frac{{5 \pi}}{{3}}) is correct because the function is increasing where its derivative is positive, which occurs in this interval.
368 Medium
368 of 402
Application +4 | -1
If variables xx and yare related by the equation x=∫0y11+9u2du,x = \int \nolimits_0^y \frac{1}{{\sqrt {1 + 9{u^2}} }}du, then d2ydx2\frac{{{d^2}y}}{{d{x^2}}} is equal to
View answer and explanation
Correct option: C
9y9y
Option C (9y9y) is correct because after differentiating the given equation twice and simplifying, we find that d2ydx2=9y\frac{{{d^2}y}}{{d{x^2}}} = 9y.
369 Medium
369 of 402
Application +4 | -1
If f(x)=x3+2x2 f′(1)+xf′′(2)+f′′′(3).f(x) = x^3 + 2x^2\,f^\prime(1) + xf^{\prime\prime}(2) + f^{\prime\prime\prime}(3). The value of f′(10)f^\prime(10) is equal to _____.
View answer and explanation
Correct option: A
218
Option A (218) is correct because it is the value of f′(10)f'(10) obtained after finding f′(1)f'(1), f′′(2)f''(2), and substituting them into the expression for f′(x)f'(x).
370 Medium
370 of 402
Application +4 | -1
If y=cos⁡23x2−sin⁡23x2y = {\cos ^2}\frac{{3x}}{2} - {\sin ^2}\frac{{3x}}{2}, then d2ydx2\frac{{{d^2}y}}{{d{x^2}}}
View answer and explanation
Correct option: C
−9y- 9y
Option C (-9y) is correct because the second derivative of y = cos(3x) is -9cos(3x), which is -9y.
371 Medium
371 of 402
Application +4 | -1
If y=sin⁡−1{x−ax−a−ax}y = {\sin ^{ - 1}}\left\{ {\sqrt {x - ax} - \sqrt {a - ax} } \right\}, then dydx\frac{{dy}}{{dx}} is equal to
View answer and explanation
Correct option: C
12x(1−x)\frac{1}{{2\sqrt {x\left( {1 - x} \right)} }}
Option C (12x(1−x)\frac{1}{{2\sqrt {x\left( {1 - x} \right)} }}) is correct because it is the derivative of the simplified inverse sine function after applying trigonometric identities.
372 Medium
372 of 402
Application +4 | -1
ddx[log⁡(tan⁡x)]\frac{d}{d x}[\log (\tan x)] is equal to
View answer and explanation
Correct option: D
2cosec⁡2x2 \operatorname{cosec} 2 x
Option D (2cosec⁡2x2 \operatorname{cosec} 2 x) is correct because differentiating and simplifying the expression leads to this result.
373 Medium
373 of 402
Analyzing +4 | -1
The derivative of f(x)=∣x∣3f\left( x \right) = {\left| x \right|^3} at x=0x = 0 is
View answer and explanation
Correct option: B
0
Option B (0) is correct because the left-hand derivative and right-hand derivative of the function at x=0 both equal 0, indicating the derivative exists and is 0.
374 Medium
374 of 402
Understanding +4 | -1
Suppose f(x)f(x) is differentiable for all xx and lim⁡h→01hf(1+h)=5\lim\limits _{h \rightarrow 0} \frac{1}{h} f(1+h)=5, then f′f^{\prime} (1) equals
View answer and explanation
Correct option: B
5
Option B (5) is correct because f′(1)=lim⁡h→0f(1+h)−f(1)hf'(1) = \lim\limits_{h \to 0} \frac{{f(1 + h) - f(1)}}{h} and given lim⁡h→01hf(1+h)=5\lim\limits _{h \rightarrow 0} \frac{1}{h} f(1+h)=5, it implies f(1)=0f(1)=0 and f′(1)=5f'(1)=5.
375 Medium
375 of 402
Application +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 dydx\frac{{dy}}{{dx}} at x=−1x = - 1
View answer and explanation
Correct option: C
−114- \frac{1}{{14}}
Option C (−114\frac{{ - 1}}{{14}}) is correct because it is the value of dydx\frac{{dy}}{{dx}} at x=−1x = -1 after solving for f(x)f(x) and differentiating y=x2f(x)y = x^2 f(x).

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376 Medium
376 of 402
Application +4 | -1
If f(3x−43x+4)=x+2f\left(\frac{3 x-4}{3 x+4}\right)=x+2 then ∫f(x)dx\int f(x) d x is equal to
View answer and explanation
Correct option: B
−83ln⁡∣(1−x)∣+23x+c-\frac{8}{3} \ln |(1-x)|+\frac{2}{3} x+c
Option B (−83ln⁡∣(1−x)∣+23x+c-\frac{8}{3} \ln |(1-x)|+\frac{2}{3} x+c) is correct because after substituting and integrating, the result matches this option.
377 Medium
377 of 402
Application +4 | -1
If f:(−1,1)→Rf:\left( { - 1,1} \right) \to R be a differentiable function with f(0)=−1f\left( 0 \right) = - 1 and f′(0)=1.f'\left( 0 \right) = 1. Let g(x)=[f(2f(x)+2)]2g(x) = {\left[ {f\left( {2f\left( x \right) + 2} \right)} \right]^2}. Then, g′(0)\mathrm{g'}\left( 0 \right) is equal to
View answer and explanation
Correct option: A
-4
Option A (-4) is correct because after applying the chain rule and substituting x=0x=0, we find that g′(0)=−4g'(0) = -4.
378 Medium
378 of 402
Analyzing +4 | -1
Let f(x+y2)=f(x)+f(y)2f\left( {\frac{{x + y}}{2}} \right) = \frac{{f(x) + f(y)}}{2} and f′(0)=a  and   f(0)=b.f'(0) = a\mathrm{ \,\,and}\,\,\,\mathrm{ }f(0) = b. Find f′′(x)f''(x) (where yy is independent of xx), when f(x)f(x) is differentiable.
View answer and explanation
Correct option: A
0
Option A (0) is correct because the second derivative of f(x) is 0, indicating that f(x) is a linear function.
379 Medium
379 of 402
Application +4 | -1
If y=sin⁡−12x⋅1−x2+sec⁡−11(1−x2)y=\sin ^{-1} 2 x \cdot \sqrt{1-x^2}+\sec ^{-1} \frac{1}{\sqrt{\left(1-x^2\right)}}, then dydx=\frac{d y}{d x}=
View answer and explanation
Correct option: C
31−x2\frac{3}{\sqrt{1-x^2}}
Option C (31−x2\frac{3}{\sqrt{1-x^2}}) is correct because after substituting x=sin⁡θx = \sin \theta and simplifying the expression, the derivative is found to be 31−x2\frac{3}{\sqrt{1-x^2}}.
380 Medium
380 of 402
Application +4 | -1
Let f(x)=2x(x−1)f(x)=2^{x(x-1)} for all x≥1x \geq 1. Then (f−1)′(4)\left(f^{-1}\right)^{\prime}(4) is (1/k)log⁡2e(1 / k) \log _2 e where the value of kk is
View answer and explanation
Correct option: D
12
Option D (12) is correct. The derivative of the inverse function is calculated and evaluated at x=4, leading to k=12.
381 Medium
381 of 402
Application +4 | -1
If y=sin⁡−1(1+x)+(1−x)2y=\sin ^{-1} \frac{\sqrt{(1+x)}+\sqrt{(1-x)}}{2}, then dydx=\frac{d y}{d x}=
View answer and explanation
Correct option: C
−12(1−x2)-\frac{1}{2 \sqrt{\left(1-x^2\right)}}
Option C (−12(1−x2)-\frac{1}{2 \sqrt{\left(1-x^2\right)}}) is correct because it is the result of simplifying the expression using trigonometric substitution and then differentiating.
382 Medium
382 of 402
Application +4 | -1
Let f:(−1,1)→Rf:(-1,1) \rightarrow R be a differentiable function with f(0)=1f(0)=1 and f′(0)=−1.f^{\prime}(0)=-1. If g(x)=(f(nf(x)−n))ng(x)=(f(n f(x)-n))^n, where nn is a natural number and g′(0)=4,g^{\prime}(0)= 4, then nn equals
View answer and explanation
Correct option: A
2
Option A is correct because solving for n in the given conditions leads to n = 2 after applying the chain rule and substituting the given values.
383 Medium
383 of 402
Analyzing +4 | -1
Let for i=1,2,3,pi(x)\mathrm{i}=1,2,3, \mathrm{p}_{\mathrm{i}}(\mathrm{x}) be a polynomial of degree 2 in x,pi′(x)\mathrm{x}, \mathrm{p}_{\mathrm{i}}^{\prime}(\mathrm{x}) and p′′′(x)\mathrm{p}_{\mathrm{\prime \prime}}{ }_{\mathrm{\prime}}(\mathrm{x}) be the first and second order derivatives of pi(x)\mathrm{p}_{\mathrm{i}}(\mathrm{x}) respectively. Let,
A(x)=[p1(x)p1′(x)p1′′(x)p2(x)p2′(x)p2′′(x)p3(x)p3′(x)p3′′(x)]A(x)=\left[\begin{array}{lll} \mathrm{p}_{1}(\mathrm{x}) & \mathrm{p}_{1}^{\prime}(\mathrm{x}) & \mathrm{p}_{1}^{\prime \prime}(\mathrm{x}) \\ \mathrm{p}_{2}(\mathrm{x}) & \mathrm{p}_{2}{ }^{\prime}(\mathrm{x}) & \mathrm{p}_{2}^{\prime \prime}(\mathrm{x}) \\ \mathrm{p}_{3}(\mathrm{x}) & \mathrm{p}_{3}{ }^{\prime}(\mathrm{x}) & \mathrm{p}_{3}{ }^{\prime \prime}(\mathrm{x}) \end{array}\right]

and B(x)=[A(x)]TA(x)\mathrm{B}(\mathrm{x})=[\mathrm{A}(\mathrm{x})]^{T} \mathrm{A}(\mathrm{x}). Then determinant of B(x)\mathrm{B}(\mathrm{x}) :
View answer and explanation
Correct option: A
is a polynomial of degree 6 in xx.
Option A is correct. The determinant of B(x) is a polynomial of degree 6 because the elements of B(x) are at most degree 4, and the determinant involves products of these elements.
384 Medium
384 of 402
Application +4 | -1
A function y=f(x)y=f(x) satisfies xdydx=y2+1,x>0&f(2)=0x \frac{d y}{d x}=\sqrt{y^2+1}, x>0 \& f(2)=0, then f(4)=f(4)=
View answer and explanation
Correct option: B
0.75
Option B (0.75) is correct because solving the differential equation and applying the initial condition f(2)=0 yields f(4)=0.75.
385 Medium
385 of 402
Analyzing +4 | -1
If z=tan⁡(y+ax)+y−ax⇒zxx−a2zyyz = \tan \left( {y + ax} \right) + \sqrt {y - ax} \Rightarrow {z_{xx}} - {a^2}{z_{yy}} is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because after finding the second-order partial derivatives and substituting them into the expression, the result is zero.
386 Medium
386 of 402
Application +4 | -1
If z=sec⁡−1(x4+y4−8x2y2x2+y2),z = {\sec ^{ - 1}}\left( {\frac{{{x^4} + {y^4} - 8{x^2}{y^2}}}{{{x^2} + {y^2}}}} \right), then x∂z∂x+y∂z∂yx\frac{{\partial z}}{{\partial x}} + y\frac{{\partial z}}{{\partial y}} is equal to
View answer and explanation
Correct option: B
2cot⁡z2\cot z
Option B (2cot⁡z2\cot z) is correct because the given function is homogeneous and Euler's theorem can be applied to find the value of the expression.
387 Medium
387 of 402
Application +4 | -1
If gg is inverse of ff and f′(x)=11+xnf^{\prime}(x)=\frac{1}{1+x^n} then g′(x)g^{\prime}(x) equals
View answer and explanation
Correct option: C
1+[g(x)]n1+[g(x)]^n
Option C (1+[g(x)]n1+[g(x)]^n) is correct because it is derived from the relationship between the derivatives of a function and its inverse.
388 Medium
388 of 402
Application +4 | -1
If f(x)=sin⁡−1(2×3x1+9x)f(x)=\sin ^{-1}\left(\frac{2 \times 3^{x}}{1+9^{x}}\right), then f′(−12)f^{\prime}\left(-\frac{1}{2}\right) equals.
View answer and explanation
Correct option: A
3log⁡e3\sqrt{3} \log _{e} \sqrt{3}
Option A (3log⁡e3\sqrt{3} \log _{e} \sqrt{3}) is correct because it is the value of the derivative of the given function at the specified point.
389 Medium
389 of 402
Application +4 | -1
ddx(tan⁡−1cos⁡x1+sin⁡x)\frac{d}{d x}\left(\tan ^{-1} \frac{\cos x}{1+\sin x}\right) 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 simplifying the trigonometric expression inside the inverse tangent, the derivative is a constant.
390 Medium
390 of 402
Application +4 | -1
If 5f(x)+3f(1x)=x+25f\left( x \right) + 3f\left( {\frac{1}{x}} \right) = x + 2 and y=xf(x),y = x\mathrm{ }f\left( x \right), then (dydx)x=1{\left( {\frac{{dy}}{{dx}}} \right)_{x = 1}} is equal to
View answer and explanation
Correct option: B
78\frac{7}{8}
Option B (78\frac{7}{8}) is correct because solving the system of equations and differentiating yields the correct result at x=1.
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