If y = sin^{-1}(frac{2x}{1 + x^2}),then find | vedclass

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(C) Given $y = \sin^{-1}(\frac{2x}{1 + x^2})$.Let $x = 	an 	heta$,then $	heta = 	an^{-1} x$.Since $x = -2$,we have $	heta = 	an^{-1}(-2)$.The ex

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$-\frac{2}{5}$

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(C) Given $y = \sin^{-1}(\frac{2x}{1 + x^2})$.Let $x = 	an 	heta$,then $	heta = 	an^{-1} x$.Since $x = -2$,we have $	heta = 	an^{-1}(-2)$.The expression becomes $y = \sin^{-1}(\frac{2 	an 	heta}{1 + 	an^2 	heta}) = \sin^{-1}(\sin 2	heta)$.For $x = -2$,$	heta = 	an^{-1}(-2)$,which lies in the interval $(-\frac{\pi}{2}, 0)$. Thus $2	heta \in (-\pi, 0)$.Since $\sin^{-1}(\sin 2	heta)$ for $2	heta \in (-\pi, -\frac{\pi}{2})$ is $-\pi - 2	heta$ and for $2	heta \in [-\frac{\pi}{2}, 0]$ is $2	heta$,we check the value: $	an^{-1}(-2) \approx -63.4^\circ$,so $2	heta \approx -126.8^\circ$.Thus,$y = -\pi - 2	heta = -\pi - 2	an^{-1} x$.Differentiating with respect to $x$:$\frac{dy}{dx} = -\frac{2}{1 + x^2}$.At $x = -2$:$\left. \frac{dy}{dx} ight|_{x = -2} = -\frac{2}{1 + (-2)^2} = -\frac{2}{1 + 4} = -\frac{2}{5}$.

If $y = \sin^{-1}(\frac{2x}{1 + x^2})$,then find the value of $\left. \frac{dy}{dx} \right|_{x = -2}$.

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