Explanation

(b)
Given, $f(x) = \sin^{- 1}\left( \frac{2x}{1 + x^{2}} \right)$
$\Rightarrow \ \ f^{'}(x) = \frac{1}{\sqrt{1 - \left( \frac{2x}{1 + x^{2}} \right)^{2}}} \times \frac{d}{dx}\left( \frac{2x}{1 + x^{2}} \right)$
$= \frac{1 + x^{2}}{\sqrt{\left( 1 + x^{2} \right)^{2}}} \times \frac{2(1 - x^{2})}{\left( 1 + x^{2} \right)^{2}}$
$= \frac{2}{1 + x^{2}} \times \frac{1 - x^{2}}{|1 - x^{2}|} = \left\{ \begin{array}{r}
\frac{2}{1 + x^{2}},\ if\ |x| < 1 \\
- \frac{2}{1 + x^{2}},\ \ if\ |x| > 1
\end{array} \right.\$
$\therefore\ \ f^{'}(x)$ does not exist for $|x| = i,\ ie,\ x = \pm 1$
Hence, $f(X)$ is differentiable on $R - \{ - 1,\ 1\}$