Let $y = \cos^{-1}\left(\frac{1 - x^{2}}{1 + x^{2}}\right)$ for every real $x$. Which of the following statements are true?

I. $y = 2\tan^{-1} x$ for $x \ge 0$.

II. $\frac{dy}{dx} = \frac{2}{1 + x^{2}}$ for every $x \ne 0$.

III. $\frac{dy}{dx} = -\frac{2}{1 + x^{2}}$ for $x < 0$.

IV. $y$ is not differentiable at $x = 0$.

Explanation

1. 01

   Put $x = \tan\theta$: $\frac{1 - x^{2}}{1 + x^{2}} = \cos 2\theta$. For $x \ge 0$, $2\theta$ lies in $[0, \pi)$, so $y = 2\tan^{-1} x$; for $x < 0$, $2\theta$ lies in $(-\pi, 0)$, so $y = -2\tan^{-1} x$.

2. 02

   So I and III are true and II is false. At 0 the slope from the right is 2 and from the left is $-2$, so IV is true.