If f(x)=sin ^{-1}left(frac{2 x}{1+x^{2}}right | vedclass

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(B) Given,$f(x)=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$.On differentiating with respect to $x$,we get:$f^{\prime}(x) = \frac{1}{\sqrt{1-\left(\fra

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$R-\{-1,1\}$

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(B) Given,$f(x)=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}ight)$.On differentiating with respect to $x$,we get:$f^{\prime}(x) = \frac{1}{\sqrt{1-\left(\frac{2 x}{1+x^{2}}ight)^{2}}} 	imes \frac{d}{d x}\left(\frac{2 x}{1+x^{2}}ight)$$f^{\prime}(x) = \frac{1+x^{2}}{\sqrt{(1-x^{2})^{2}}} 	imes \frac{2(1-x^{2})}{(1+x^{2})^{2}}$$f^{\prime}(x) = \frac{2}{1+x^{2}} 	imes \frac{1-x^{2}}{|1-x^{2}|}$This simplifies to:$f^{\prime}(x) = \begin{cases} \frac{2}{1+x^{2}}, & 	ext{if } |x|  1 \end{cases}$At $x = 1$ and $x = -1$,the derivative does not exist because the left-hand derivative and right-hand derivative are not equal.Therefore,$f(x)$ is differentiable on $R - \{-1, 1\}$.

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

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