Find frac{dy}{dx} if y = sin^{-1}left(frac{1- | vedclass

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(A) Let $x = 	an 	heta$,then $	heta = 	an^{-1} x$. Since $0 < x < 1$,we have $0 < 	heta < \frac{\pi}{4}$.The expression becomes $y = \sin^{-1}\le

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

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(A) Let $x = 	an 	heta$,then $	heta = 	an^{-1} x$. Since $0 The expression becomes $y = \sin^{-1}\left(\frac{1-	an^2 	heta}{1+	an^2 	heta}ight)$.Using the trigonometric identity $\cos 2	heta = \frac{1-	an^2 	heta}{1+	an^2 	heta}$,we get $y = \sin^{-1}(\cos 2	heta)$.Since $\cos 2	heta = \sin\left(\frac{\pi}{2} - 2	hetaight)$,we have $y = \sin^{-1}\left(\sin\left(\frac{\pi}{2} - 2	hetaight)ight)$.Given $0 Thus,$y = \frac{\pi}{2} - 2	heta = \frac{\pi}{2} - 2	an^{-1} x$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2}ight) - 2\frac{d}{dx}(	an^{-1} x) = 0 - 2\left(\frac{1}{1+x^2}ight) = \frac{-2}{1+x^2}$.

Find $\frac{dy}{dx}$ if $y = \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right)$,where $0 < x < 1$.

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