Differentiate the following w.r.t. $\tan^{-1} \left(\frac{2x}{1-x^2}\right)$ Differentiate : $$ \tan^{-1}  \left(\frac {\sqrt {1+x^2}-1}x\right) \quad w.r.t.\quad \tan^{-1} \left(\frac{2x}{1-x^2}\right) $$       

Let $y=\tan^{-1}  \left(\dfrac {\sqrt {1+x^2}-1}x\right)$
and let $u= \tan^{-1} \left(\dfrac{2x}{1-x^2}\right)$.
We want to find $dy/du$. Note that:
$$
\dfrac{dy}{dx} = \dfrac{1}{2(1+x^2)}
$$
similarly for $u$, we obtain:
$$
\dfrac{du}{dx} = \dfrac{2}{1+x^2} \iff \dfrac{dx}{du} = \dfrac{1+x^2}{2}
$$
Hence, by Chain Rule, we obtain:
$$
\dfrac{dy}{du} = \dfrac{dy}{dx} \cdot \dfrac{dx}{du} = \dfrac{1}{2(1+x^2)} \cdot \dfrac{1+x^2}{2} = \dfrac{1}{4}
$$
