What is $\frac{dy}{dx}$ for $y = \sin^{-1}\left(\frac{2x}{1 + x^{2}}\right)$, $x > 1$?

Explanation

1. 01

   Put $x = \tan\theta$ with $\frac{\pi}{4} < \theta < \frac{\pi}{2}$: $\frac{2x}{1 + x^{2}} = \sin 2\theta$, and $2\theta$ lies in $\left(\frac{\pi}{2}, \pi\right)$.

2. 02

   There $\sin^{-1}(\sin 2\theta) = \pi - 2\theta$, so $y = \pi - 2\tan^{-1} x$ and $\frac{dy}{dx} = -\frac{2}{1 + x^{2}}$.