If y = sin^{-1}left(frac{2x}{1+x^2}right) + s | vedclass

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(C) Let $x = 	an 	heta$. Then $	heta = 	an^{-1} x$. For $x = \sqrt{3}$,$	heta = \frac{\pi}{3}$. The expression becomes $y = \sin^{-1}(\sin 2	het

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(C) Let $x = 	an 	heta$. Then $	heta = 	an^{-1} x$. For $x = \sqrt{3}$,$	heta = \frac{\pi}{3}$. The expression becomes $y = \sin^{-1}(\sin 2	heta) + \sec^{-1}(\sec 2	heta)$. Since $x = \sqrt{3} > 1$,we must be careful with the range. For $x > 1$,$\sin^{-1}(\frac{2x}{1+x^2}) = \pi - 2	an^{-1} x$ and $\sec^{-1}(\frac{1+x^2}{1-x^2}) = \pi + 2	an^{-1} x$ is not correct here. Let's use the derivative directly: $\frac{d}{dx}(\sin^{-1}(\frac{2x}{1+x^2})) = \frac{2}{1+x^2}$ for $|x|  1$. $\frac{d}{dx}(\sec^{-1}(\frac{1+x^2}{1-x^2})) = \frac{d}{dx}(\cos^{-1}(\frac{1-x^2}{1+x^2})) = -\frac{2}{1+x^2}$ for $x > 1$. Thus,for $x > 1$,$\frac{dy}{dx} = -\frac{2}{1+x^2} - \frac{2}{1+x^2} = -\frac{4}{1+x^2}$. At $x = \sqrt{3}$,$\frac{dy}{dx} = -\frac{4}{1+3} = -\frac{4}{4} = -1$. However,checking the standard identity $\sec^{-1}(\frac{1+x^2}{1-x^2}) = \cos^{-1}(\frac{1-x^2}{1+x^2})$,for $x > 1$,this is $2	an^{-1} x - \pi$. So $y = (\pi - 2	an^{-1} x) + (2	an^{-1} x - \pi) = 0$. Therefore,$\frac{dy}{dx} = 0$.

If $y = \sin^{-1}\left(\frac{2x}{1+x^2}\right) + \sec^{-1}\left(\frac{1+x^2}{1-x^2}\right)$,then the value of $\frac{dy}{dx}$ at $x = \sqrt{3}$ is

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