The function y = sin^{-1}left(frac{2x}{1 + x^ | vedclass

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(B) Let $f(x) = \sin^{-1}\left(\frac{2x}{1 + x^2}ight)$.We know that $\frac{2x}{1 + x^2} = \sin(2	an^{-1}x)$.Thus,$y = \sin^{-1}(\sin(2	an^{-1}x))

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$x = 1, -1$

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(B) Let $f(x) = \sin^{-1}\left(\frac{2x}{1 + x^2}ight)$.We know that $\frac{2x}{1 + x^2} = \sin(2	an^{-1}x)$.Thus,$y = \sin^{-1}(\sin(2	an^{-1}x))$.Case $1$: If $|x| \le 1$,then $-\frac{\pi}{2} \le 2	an^{-1}x \le \frac{\pi}{2}$,so $y = 2	an^{-1}x$. The derivative is $y' = \frac{2}{1 + x^2}$.Case $2$: If $x > 1$,then $2	an^{-1}x > \frac{\pi}{2}$,so $y = \pi - 2	an^{-1}x$. The derivative is $y' = -\frac{2}{1 + x^2}$.Case $3$: If $x At $x = 1$,the left-hand derivative is $\frac{2}{1 + 1^2} = 1$ and the right-hand derivative is $-\frac{2}{1 + 1^2} = -1$.At $x = -1$,the left-hand derivative is $-\frac{2}{1 + (-1)^2} = -1$ and the right-hand derivative is $\frac{2}{1 + (-1)^2} = 1$.Since the left-hand and right-hand derivatives are not equal at $x = 1$ and $x = -1$,the function is not differentiable at these points.

The function $y = \sin^{-1}\left(\frac{2x}{1 + x^2}\right)$ is not differentiable for

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