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JEE MATHEMATICS MCQ: DIFFERENTIATION

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If f(x)=sin⁡−1(3x−4x3)f(x)=\sin ^{-1}\left(3 x-4 x^3\right) for all x∈[−1,1]x \in[-1,1] then
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Correct option: C
f'(x) = \left\{ {\begin{array}{llllllllllllllllllll} {\frac{3}{{\sqrt {1 - {x^2}} }},}&{\mathrm{ if}\,\,\mathrm{ } - \frac{1}{2} < x < \frac{1}{2}}\\ {\frac{{ - 3}}{{\sqrt {1 - {x^2}} }},}&{\mathrm{ if }\,\,\frac{1}{2} < x < 1\,\,\,\mathrm{ or},\,\,\, - 1 < x < - \frac{1}{f'(x) = \left\{ {\begin{array}{llllllllllllllllllll} {\frac{3}{{\sqrt {1 - {x^2}} }},}&{\mathrm{ if}\,\,\mathrm{ } - \frac{1}{2} < x < \frac{1}{2}}\\ {\frac{{ - 3}}{{\sqrt {1 - {x^2}} }},}&{\mathrm{ if }\,\,\frac{1}{2} < x < 1\,\,\,\mathrm{ or},\,\,\, - 1 < x < - \frac{1}{
Option C is correct because the derivative of arcsin(3x-4x3) is piecewise, being 3/sqrt(1-x2) for |x|<1/2 and -3/sqrt(1-x2) for 1/2<|x|<1 due to the properties of arcsin.

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If f(x)=∣x∣∣tan⁡x∣f(x)=|x|^{|\tan x|}, then f′(−π6)f^{\prime}\left(\frac{-\pi}{6}\right) is equal to Let xx and yy be two real numbers such that x>0x>0 and xy=1x y=1. The minimum value of x+yx+y is Let f(x)=[x]sin⁡(πx)f(x)=[x] \sin (\pi x) where [x][x] denotes the integer part of xx. Then at x=k,kx=k, k being an integer, the left derivative of f(x)f(x) is If y=sec⁡−1[cosecx]+cosec−1[sec⁡x]+sin⁡−1[cos⁡x]+cos⁡−1[sin⁡x],y = {\sec ^{ - 1}}\left[ {\mathrm{cosec}x} \right] + \mathrm{cose}{\mathrm{c}^{ - 1}}\left[ {\sec x} \right] + {\sin ^{ - 1}}[\cos x] + {\cos ^{ - 1}}[\sin x], then dydx\frac{{dy}}{{dx}} is equal to ddxsin⁡−1(3x−4x3)\frac{d}{d x} \sin ^{-1}\left(3 x-4 x^{3}\right) is equal to If z=y+f(v),z = y + f\left( v \right), where v=(xy),v = \left( {\frac{x}{y}} \right), then v∂z∂x+∂z∂yv\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}} is
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