Question
If $u = \sin^{-1}\left(\frac{2x}{1 + x^2}\right)$ and $v = \tan^{-1}\left(\frac{2x}{1 - x^2}\right)$, then $\frac{du}{dv}$ is:
If $u = \sin^{-1}\left(\frac{2x}{1 + x^2}\right)$ and $v = \tan^{-1}\left(\frac{2x}{1 - x^2}\right)$, then $\frac{du}{dv}$ is: