The derivative of $\tan ^{-1}\left(\frac{1+x}{1-x}\right)$ is

JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1

The derivative of

\tan ^{-1}\left(\frac{1+x}{1-x}\right)

View answer and explanation

Correct option: B

\frac{1}{1+x^2}

Option B (

\frac{1}{1+x^2}

) is correct because after substituting

x = \tan \theta

and simplifying the expression, the derivative is found to be

\frac{1}{1+x^2}

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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}}