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

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(C) Given $y = \sin^{-1}\left( \frac{1 - x^2}{1 + x^2} ight)$.Let $x = 	an 	heta$,then $	heta = 	an^{-1} x$.Substituting this into the expressio

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$\pm \frac{2}{1 + x^2}$

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(C) Given $y = \sin^{-1}\left( \frac{1 - x^2}{1 + x^2} ight)$.Let $x = 	an 	heta$,then $	heta = 	an^{-1} x$.Substituting this into the expression,we get $y = \sin^{-1}(\cos 2	heta)$.Since $\cos 2	heta = \sin\left( \frac{\pi}{2} - 2	heta ight)$ or $\sin\left( 2	heta - \frac{\pi}{2} ight)$,we have $y = \frac{\pi}{2} \pm 2	heta$.Substituting $	heta = 	an^{-1} x$,we get $y = \frac{\pi}{2} \pm 2 	an^{-1} x$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = \pm 2 \cdot \frac{1}{1 + x^2} = \pm \frac{2}{1 + x^2}$.

If $y = \sin^{-1}\left( \frac{1 - x^2}{1 + x^2} \right)$,then $\frac{dy}{dx}$ equals

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