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

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

The correct answer is Option (2) → $\frac{1}{(1+tan^{-1}x)^2}$

$y=\frac{\tan^{-1}x}{1+\tan^{-1}x}$, $z=\tan^{-1}x$

$\frac{dy}{dx}=\frac{\frac{1}{1+x^2}(1+\tan^{-1}x)-\frac{1}{1+x^2}\tan^{-1}x}{(1+\tan^{-1}x)^2}$,   $\frac{dz}{dx}=\frac{1}{1+x^2}$

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

$⇒\frac{dy}{dz}=\frac{1}{(1+tan^{-1}x)^2}$