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JEE MATHEMATICS MCQ: DIFFERENTIATION

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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
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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.

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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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