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

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(C) Let $x = \cos 	heta$. Then $	heta = \cos^{-1} x$.Using the identities $1 + \cos 	heta = 2\cos^2(\frac{	heta}{2})$ and $1 - \cos 	heta = 2\sin

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

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(C) Let $x = \cos 	heta$. Then $	heta = \cos^{-1} x$.Using the identities $1 + \cos 	heta = 2\cos^2(\frac{	heta}{2})$ and $1 - \cos 	heta = 2\sin^2(\frac{	heta}{2})$,we have:$y = \sin^{-1} \left[ \frac{\sqrt{2}\cos(\frac{	heta}{2}) + \sqrt{2}\sin(\frac{	heta}{2})}{2} ight]$$y = \sin^{-1} \left[ \frac{1}{\sqrt{2}}\cos(\frac{	heta}{2}) + \frac{1}{\sqrt{2}}\sin(\frac{	heta}{2}) ight]$Since $\sin(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$,we can write:$y = \sin^{-1} \left[ \sin(\frac{\pi}{4})\cos(\frac{	heta}{2}) + \cos(\frac{\pi}{4})\sin(\frac{	heta}{2}) ight]$$y = \sin^{-1} \left[ \sin(\frac{	heta}{2} + \frac{\pi}{4}) ight]$$y = \frac{	heta}{2} + \frac{\pi}{4}$Substituting $	heta = \cos^{-1} x$:$y = \frac{1}{2}\cos^{-1} x + \frac{\pi}{4}$Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{1}{2} \left( -\frac{1}{\sqrt{1-x^2}} ight) = -\frac{1}{2\sqrt{1-x^2}}$.

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

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