What is $\frac{dy}{dx}$ for $y = \tan^{-1}\sqrt{\frac{1 - \cos x}{1 + \cos x}}$, $0 < x < \pi$?

Explanation

1. 01

   $\frac{1 - \cos x}{1 + \cos x} = \frac{2\sin^{2}(x/2)}{2\cos^{2}(x/2)} = \tan^{2}\frac{x}{2}$, and $\tan\frac{x}{2} > 0$ for $0 < x < \pi$.

2. 02

   So $y = \tan^{-1}\left(\tan\frac{x}{2}\right) = \frac{x}{2}$, since $\frac{x}{2}$ lies in $\left(0, \frac{\pi}{2}\right)$: $\frac{dy}{dx} = \frac{1}{2}$.