If $f(x)=2 \tan ^{-1} x+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)$, then

JEE MATHEMATICS MCQ: DIFFERENTIATION

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f(x)=2 \tan ^{-1} x+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)

, then

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Correct option: C

f^{\prime}(x)=0

for all

x < 0

Option D (All the above) is correct because all three given statements are true based on the given functions and their derivatives.

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f(x)=|x|^{|\tan x|}

f^{\prime}\left(\frac{-\pi}{6}\right)

x

y

x>0

x y=1

x+y

f(x)=[x] \sin (\pi x)

[x]

x

x=k, k

f(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],

\frac{{dy}}{{dx}}

\frac{d}{d x} \sin ^{-1}\left(3 x-4 x^{3}\right)

z = y + f\left( v \right),

v = \left( {\frac{x}{y}} \right),

v\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}}