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  4. If f(x) = sin -1 ((2x/1+x2)) , then f(x) is differentiable on

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

MHT CETMHT CET 2007

Solution:

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