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

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(A) Given $y = \sin^{-1}\left(\frac{2x}{1+x^2}ight)$.Let $x = 	an 	heta$,then $	heta = 	an^{-1} x$.Substituting $x = 	an 	heta$ in the express

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

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(A) Given $y = \sin^{-1}\left(\frac{2x}{1+x^2}ight)$.Let $x = 	an 	heta$,then $	heta = 	an^{-1} x$.Substituting $x = 	an 	heta$ in the expression,we get:$y = \sin^{-1}\left(\frac{2 	an 	heta}{1 + 	an^2 	heta}ight)$Using the trigonometric identity $\sin 2	heta = \frac{2 	an 	heta}{1 + 	an^2 	heta}$,we have:$y = \sin^{-1}(\sin 2	heta)$Since $|x| Substituting back $	heta = 	an^{-1} x$:$y = 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 $y = \sin^{-1}\left(\frac{2x}{1+x^2}\right)$ where $|x| < 1$.

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