If y=sin ^{-1}left[x sqrt{1-x^2}-sqrt{x} sqrt | vedclass

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(A) Given $y = \sin^{-1}(x \sqrt{1-x^2} - \sqrt{x} \sqrt{1-x})$.Let $x = \sin \alpha$ and $\sqrt{x} = \sin \beta$. Then $\sqrt{1-x^2} = \cos \alpha$ a

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

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(A) Given $y = \sin^{-1}(x \sqrt{1-x^2} - \sqrt{x} \sqrt{1-x})$.Let $x = \sin \alpha$ and $\sqrt{x} = \sin \beta$. Then $\sqrt{1-x^2} = \cos \alpha$ and $\sqrt{1-x} = \cos \beta$.Substituting these into the expression,we get $y = \sin^{-1}(\sin \alpha \cos \beta - \cos \alpha \sin \beta)$.Using the identity $\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$,we have $y = \sin^{-1}(\sin(\alpha - \beta)) = \alpha - \beta$.Thus,$y = \sin^{-1} x - \sin^{-1} \sqrt{x}$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(\sin^{-1} x) - \frac{d}{dx}(\sin^{-1} \sqrt{x})$.$\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-(\sqrt{x})^2}} \cdot \frac{d}{dx}(\sqrt{x})$.$\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-x}} \cdot \frac{1}{2\sqrt{x}}$.$\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} - \frac{1}{2\sqrt{x(1-x)}} = \frac{1}{\sqrt{1-x^2}} - \frac{1}{2\sqrt{x-x^2}}$.

If $y=\sin ^{-1}\left[x \sqrt{1-x^2}-\sqrt{x} \sqrt{1-x}\right]$ and $0 < x < 1$,then $\frac{d y}{d x}$ is equal to

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