(C) Let $u = \tan^{-1}x$ and $y = \frac{u}{1+u}$.We need to find $\frac{dy}{du}$.Using the quotient rule for differentiation,$\frac{dy}{du} = \frac{d}

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$\frac{1}{{{(1 + \tan^{-1}x)}^2}}$

(C) Let $u = \tan^{-1}x$ and $y = \frac{u}{1+u}$.We need to find $\frac{dy}{du}$.Using the quotient rule for differentiation,$\frac{dy}{du} = \frac{d}{du} \left( \frac{u}{1+u} \right)$.$\frac{dy}{du} = \frac{(1+u)(1) - u(1)}{(1+u)^2}$.$\frac{dy}{du} = \frac{1+u-u}{(1+u)^2} = \frac{1}{(1+u)^2}$.Substituting $u = \tan^{-1}x$ back,we get $\frac{dy}{du} = \frac{1}{{(1 + \tan^{-1}x)}^2}$.

Class 12 Mathematics Continuity and Differentiability Derivative at a point, Standard differentiation

Medium

The differential coefficient of $\frac{{\tan^{-1}x}}{{1 + \tan^{-1}x}}$ with respect to $\tan^{-1}x$ is

A
$\frac{1}{{1 + \tan^{-1}x}}$
B
$\frac{{-1}}{{1 + \tan^{-1}x}}$
C
$\frac{1}{{{(1 + \tan^{-1}x)}^2}}$
D
$\frac{{-1}}{{2{(1 + \tan^{-1}x)}^2}}$

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Derivative at a point, Standard differentiation

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The differential coefficient of $\frac{{\tan^{-1}x}}{{1 + \tan^{-1}x}}$ with respect to $\tan^{-1}x$ is

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