If \(f(x)=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\), then \(f(x)\) is
differentiable on

Given, \(f(x)=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\)
On differentiating w.r.t. \(x\), we get
\(\begin{aligned} f^{\prime}(x) &=\frac{1}{\sqrt{1-\left(\frac{2 x}{1+x^{2}}\right)^{2}} \times \frac{d}{d x}\left(\frac{2 x}{1+x^{2}}\right)} \\ &=\frac{1+x^{2}}{\sqrt{\left(1-x^{2}\right)^{2}}} \times \frac{2\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}} \end{aligned}\)
\(=\frac{2}{1+x^{2}} \times \frac{1-x^{2}}{\left|1-x^{2}\right|}\)
\(=\left\{\begin{array}{l}\frac{2}{1+x^{2}}, \text { if }|x| < 1 \\ -\frac{2}{1+x^{2}}, \text { if }|x|>1\end{array}\right.\)
\(\therefore f^{\prime}(x)\) does not exist for \(|x|=1, i e, x=\pm 1\)
Hence, \(f(x)\) is differentiable on \(R-\{-1,1\}\).