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JEE MATHEMATICS MCQ: DIFFERENTIATION

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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
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Correct option: D
−2(x2−x−1(x2+1)2)⋅sin⁡(2x−1x2+1)2-2\left(\frac{x^2-x-1}{\left(x^2+1\right)^2}\right) \cdot \sin \left(\frac{2 x-1}{x^2+1}\right)^2
Option D (−2(x2−x−1(x2+1)2)⋅sin⁡(2x−1x2+1)2-2(\frac{x^2-x-1}{(x^2+1)^2}) \cdot \sin (\frac{2 x-1}{x^2+1})^2) is correct because it is the result of applying the chain rule.

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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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