The derivative of y = sin^{-1}left(frac{sqrt{ | vedclass

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(D) $y = \sin^{-1}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{2}ight)$ Let $x = \cos 2	heta$,then $2	heta = \cos^{-1} x$ or $	heta = \frac{1}{2} \cos^{-

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

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(D) $y = \sin^{-1}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{2}ight)$ Let $x = \cos 2	heta$,then $2	heta = \cos^{-1} x$ or $	heta = \frac{1}{2} \cos^{-1} x$. Substituting $x$ in the expression: $y = \sin^{-1}\left(\frac{\sqrt{1+\cos 2	heta} - \sqrt{1-\cos 2	heta}}{2}ight)$ Using trigonometric identities $1+\cos 2	heta = 2\cos^2 	heta$ and $1-\cos 2	heta = 2\sin^2 	heta$: $y = \sin^{-1}\left(\frac{\sqrt{2}\cos 	heta - \sqrt{2}\sin 	heta}{2}ight) = \sin^{-1}\left(\frac{\cos 	heta - \sin 	heta}{\sqrt{2}}ight)$ $y = \sin^{-1}\left(\frac{1}{\sqrt{2}}\cos 	heta - \frac{1}{\sqrt{2}}\sin 	hetaight)$ $y = \sin^{-1}\left(\sin \frac{\pi}{4} \cos 	heta - \cos \frac{\pi}{4} \sin 	hetaight)$ $y = \sin^{-1}\left(\sin\left(\frac{\pi}{4} - 	hetaight)ight) = \frac{\pi}{4} - 	heta$ Substituting $	heta = \frac{1}{2} \cos^{-1} x$: $y = \frac{\pi}{4} - \frac{1}{2} \cos^{-1} x$ Differentiating with respect to $x$: $\frac{dy}{dx} = 0 - \frac{1}{2} \left(-\frac{1}{\sqrt{1-x^2}}ight) = \frac{1}{2\sqrt{1-x^2}}$ Thus,option $D$ is correct.

The derivative of $y = \sin^{-1}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{2}\right)$ is

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