If y = sin^{-1} left[ frac{sqrt{1+x} + sqrt{1 | vedclass

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(A) Let $x = \cos(2	heta)$,so $2	heta = \cos^{-1}(x)$ and $	heta = \frac{1}{2} \cos^{-1}(x)$.Then $\sqrt{1+x} = \sqrt{1+\cos(2	heta)} = \sqrt{2\co

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

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(A) Let $x = \cos(2	heta)$,so $2	heta = \cos^{-1}(x)$ and $	heta = \frac{1}{2} \cos^{-1}(x)$.Then $\sqrt{1+x} = \sqrt{1+\cos(2	heta)} = \sqrt{2\cos^2(	heta)} = \sqrt{2} \cos(	heta)$ and $\sqrt{1-x} = \sqrt{1-\cos(2	heta)} = \sqrt{2\sin^2(	heta)} = \sqrt{2} \sin(	heta)$.Substituting these into the expression for $y$:$y = \sin^{-1} \left[ \frac{\sqrt{2}\cos(	heta) + \sqrt{2}\sin(	heta)}{2} ight] = \sin^{-1} \left[ \frac{1}{\sqrt{2}} \cos(	heta) + \frac{1}{\sqrt{2}} \sin(	heta) ight]$.Using the identity $\sin(\frac{\pi}{4} + 	heta) = \sin(\frac{\pi}{4}) \cos(	heta) + \cos(\frac{\pi}{4}) \sin(	heta) = \frac{1}{\sqrt{2}} \cos(	heta) + \frac{1}{\sqrt{2}} \sin(	heta)$.So,$y = \sin^{-1} [\sin(\frac{\pi}{4} + 	heta)] = \frac{\pi}{4} + 	heta = \frac{\pi}{4} + \frac{1}{2} \cos^{-1}(x)$.Differentiating with respect to $x$:$\frac{dy}{dx} = 0 + \frac{1}{2} \left( -\frac{1}{\sqrt{1-x^2}} ight) = -\frac{1}{2\sqrt{1-x^2}}$.

If $y = \sin^{-1} \left[ \frac{\sqrt{1+x} + \sqrt{1-x}}{2} \right]$,then $\frac{dy}{dx} = $

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