If y = frac{sin^{-1} x}{sqrt{1 - x^2}},then ( | vedclass

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(B) Given $y = \frac{\sin^{-1} x}{\sqrt{1 - x^2}}$.Applying the quotient rule for differentiation,$\frac{dy}{dx} = \frac{\sqrt{1 - x^2} \cdot \frac{d}

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$1 + xy$

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(B) Given $y = \frac{\sin^{-1} x}{\sqrt{1 - x^2}}$.Applying the quotient rule for differentiation,$\frac{dy}{dx} = \frac{\sqrt{1 - x^2} \cdot \frac{d}{dx}(\sin^{-1} x) - \sin^{-1} x \cdot \frac{d}{dx}(\sqrt{1 - x^2})}{1 - x^2}$.Since $\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}}$ and $\frac{d}{dx}(\sqrt{1 - x^2}) = \frac{1}{2\sqrt{1 - x^2}} \cdot (-2x) = \frac{-x}{\sqrt{1 - x^2}}$,we have:$\frac{dy}{dx} = \frac{\sqrt{1 - x^2} \cdot \frac{1}{\sqrt{1 - x^2}} - \sin^{-1} x \cdot \left( \frac{-x}{\sqrt{1 - x^2}} \right)}{1 - x^2}$.$\frac{dy}{dx} = \frac{1 + \frac{x \sin^{-1} x}{\sqrt{1 - x^2}}}{1 - x^2}$.Multiplying both sides by $(1 - x^2)$,we get:$(1 - x^2)\frac{dy}{dx} = 1 + x \left( \frac{\sin^{-1} x}{\sqrt{1 - x^2}} \right)$.Since $y = \frac{\sin^{-1} x}{\sqrt{1 - x^2}}$,we substitute $y$ into the expression:$(1 - x^2)\frac{dy}{dx} = 1 + xy$.

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

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