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

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(C) Let $x = \sin A$ and $\sqrt{x} = \sin B$. Then $A = \sin^{-1} x$ and $B = \sin^{-1} \sqrt{x}$. The expression becomes $y = \sin^{-1}(\sin A \cos B

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

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(C) Let $x = \sin A$ and $\sqrt{x} = \sin B$. Then $A = \sin^{-1} x$ and $B = \sin^{-1} \sqrt{x}$. The expression becomes $y = \sin^{-1}(\sin A \cos B + \sin B \cos A)$. Using the identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$,we get $y = \sin^{-1}(\sin(A + B)) = A + B$. Substituting back,$y = \sin^{-1} x + \sin^{-1} \sqrt{x}$. Differentiating with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(\sin^{-1} x) + \frac{d}{dx}(\sin^{-1} \sqrt{x})$. $\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} + \frac{1}{\sqrt{1 - (\sqrt{x})^2}} \cdot \frac{d}{dx}(\sqrt{x})$. $\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} + \frac{1}{\sqrt{1 - x}} \cdot \frac{1}{2\sqrt{x}}$. $\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} + \frac{1}{2\sqrt{x - x^2}}$.

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

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