If y = 2sin^{-1} sqrt{1-x} + sin^{-1} (2sqrt{ | vedclass

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(D) Let $\sqrt{x} = \sin 	heta$,which implies $x = \sin^2 	heta$. Since $0 < x < \frac{1}{2}$,we have $0 < \sin^2 	heta < \frac{1}{2}$,so $0 < \sin

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(D) Let $\sqrt{x} = \sin 	heta$,which implies $x = \sin^2 	heta$. Since $0 Then,$\sqrt{1-x} = \sqrt{1-\sin^2 	heta} = \cos 	heta$. Also,$2\sqrt{x(1-x)} = 2\sin 	heta \cos 	heta = \sin 2	heta$. Substituting these into the expression for $y$: $y = 2\sin^{-1}(\cos 	heta) + \sin^{-1}(\sin 2	heta)$. Since $0 Also,$\sin^{-1}(\cos 	heta) = \sin^{-1}(\sin(\frac{\pi}{2} - 	heta)) = \frac{\pi}{2} - 	heta$. Thus,$y = 2(\frac{\pi}{2} - 	heta) + 2	heta = \pi - 2	heta + 2	heta = \pi$. Since $y = \pi$ is a constant,$\frac{dy}{dx} = 0$.

If $y = 2\sin^{-1} \sqrt{1-x} + \sin^{-1} (2\sqrt{x(1-x)})$ for $0 < x < \frac{1}{2}$,then $\frac{dy}{dx}$ equals-

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