Find frac{dy}{dx} for y = cos^{-1}left(frac{1 | vedclass

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(A) Given $y = \cos^{-1}\left(\frac{1-x^2}{1+x^2}ight)$.Let $x = 	an 	heta$,where $	heta = 	an^{-1} x$. Since $0 < x < 1$,we have $0 < 	heta <

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

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(A) Given $y = \cos^{-1}\left(\frac{1-x^2}{1+x^2}ight)$.Let $x = 	an 	heta$,where $	heta = 	an^{-1} x$. Since $0 Substituting $x = 	an 	heta$ into the expression:$y = \cos^{-1}\left(\frac{1-	an^2 	heta}{1+	an^2 	heta}ight)$Using the trigonometric identity $\cos 2	heta = \frac{1-	an^2 	heta}{1+	an^2 	heta}$:$y = \cos^{-1}(\cos 2	heta)$Since $0 Thus,$y = 2	heta = 2 	an^{-1} x$.Differentiating with respect to $x$:$\frac{dy}{dx} = 2 \cdot \frac{d}{dx}(	an^{-1} x)$$\frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2} = \frac{2}{1+x^2}$.

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

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