Find frac{dy}{dx} for y = cos^{-1}left(frac{2 | vedclass

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(A) Given $y = \cos^{-1}\left(\frac{2x}{1+x^2}ight)$.Let $x = 	an 	heta$,where $	heta = 	an^{-1} x$. Since $-1 < x < 1$,we have $-\frac{\pi}{4}

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

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(A) Given $y = \cos^{-1}\left(\frac{2x}{1+x^2}ight)$.Let $x = 	an 	heta$,where $	heta = 	an^{-1} x$. Since $-1 Then $\frac{2x}{1+x^2} = \frac{2 	an 	heta}{1 + 	an^2 	heta} = \sin(2	heta)$.So,$y = \cos^{-1}(\sin(2	heta)) = \cos^{-1}(\cos(\frac{\pi}{2} - 2	heta))$.Since $-\frac{\pi}{4} Thus,$y = \frac{\pi}{2} - 2	heta = \frac{\pi}{2} - 2 	an^{-1} x$.Differentiating with respect to $x$:$\frac{dy}{dx} = 0 - 2 \cdot \frac{1}{1+x^2} = \frac{-2}{1+x^2}$.

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

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