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

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(D) Let $x = \sin^2 	heta$ and $\sqrt{x} = \sin \alpha$. Then $x = \sin^2 \alpha$ and $\sqrt{x} = \sin \alpha$. Substituting these into the expressio

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(D) Let $x = \sin^2 	heta$ and $\sqrt{x} = \sin \alpha$. Then $x = \sin^2 \alpha$ and $\sqrt{x} = \sin \alpha$. Substituting these into the expression: $y = \sin^{-1} (\sin^2 \alpha \sqrt{1 - \sin^2 \alpha} + \sin \alpha \sqrt{1 - \sin^4 \alpha})$ This substitution is complex. Let's use $x = \sin^2 	heta$ and $\sqrt{x} = \sin \phi$. Actually,let $x = \sin^2 \alpha$ and $\sqrt{x} = \sin \beta$. Then $y = \sin^{-1} (\sin^2 \alpha \cos \beta + \sin \beta \cos \alpha)$ is not standard. Let $x = \sin^2 	heta$ and $\sqrt{x} = \sin \phi$. Let $x = \sin^2 A$ and $\sqrt{x} = \sin B$. Consider $y = \sin^{-1} (x \sqrt{1 - (\sqrt{x})^2} + \sqrt{x} \sqrt{1 - x^2})$. Let $x = \sin A$ and $\sqrt{x} = \sin B$. Then $y = \sin^{-1} (\sin A \cos B + \sin B \cos A) = \sin^{-1} (\sin(A + B)) = A + B$. Since $x = \sin A$,$A = \sin^{-1} x$. Since $\sqrt{x} = \sin B$,$B = \sin^{-1} \sqrt{x}$. So,$y = \sin^{-1} x + \sin^{-1} \sqrt{x}$. Differentiating with respect to $x$: $\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} + \frac{1}{\sqrt{1 - (\sqrt{x})^2}} \cdot \frac{d}{dx}(\sqrt{x})$ $\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} + \frac{1}{\sqrt{1 - x}} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{\sqrt{1 - x^2}} + \frac{1}{2\sqrt{x(1 - x)}}$. Comparing this with $\frac{dy}{dx} = \frac{1}{2\sqrt{x(1 - x)}} + p$,we get $p = \frac{1}{\sqrt{1 - x^2}}$.

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

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