If y=frac{x sin ^{-1} x}{sqrt{1-x^2}}+log sqr | vedclass

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(B) Given $y = \frac{x \sin^{-1} x}{\sqrt{1-x^2}} + \log \sqrt{1-x^2}$.First,simplify the logarithmic term: $\log \sqrt{1-x^2} = \frac{1}{2} \log(1-x^

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$\frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}}$

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(B) Given $y = \frac{x \sin^{-1} x}{\sqrt{1-x^2}} + \log \sqrt{1-x^2}$.First,simplify the logarithmic term: $\log \sqrt{1-x^2} = \frac{1}{2} \log(1-x^2)$.Now,differentiate $y$ with respect to $x$ using the quotient rule for the first term and the chain rule for the second term:$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{x \sin^{-1} x}{\sqrt{1-x^2}} \right) + \frac{d}{dx} \left( \frac{1}{2} \log(1-x^2) \right)$.Using the quotient rule $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}$ where $u = x \sin^{-1} x$ and $v = \sqrt{1-x^2}$:$u' = \sin^{-1} x + \frac{x}{\sqrt{1-x^2}}$ and $v' = \frac{-x}{\sqrt{1-x^2}}$.$\frac{d}{dx} \left( \frac{x \sin^{-1} x}{\sqrt{1-x^2}} \right) = \frac{(\sin^{-1} x + \frac{x}{\sqrt{1-x^2}})\sqrt{1-x^2} - (x \sin^{-1} x)(\frac{-x}{\sqrt{1-x^2}})}{1-x^2} = \frac{(1-x^2)\sin^{-1} x + x + \frac{x^2 \sin^{-1} x}{\sqrt{1-x^2}}}{1-x^2} = \frac{\sin^{-1} x + \frac{x^2 \sin^{-1} x}{\sqrt{1-x^2}} + x}{1-x^2}$.Now,differentiate the second term: $\frac{d}{dx} (\frac{1}{2} \log(1-x^2)) = \frac{1}{2} \cdot \frac{1}{1-x^2} \cdot (-2x) = \frac{-x}{1-x^2}$.Adding both parts: $\frac{dy}{dx} = \frac{\sin^{-1} x + \frac{x^2 \sin^{-1} x}{\sqrt{1-x^2}} + x - x}{1-x^2} = \frac{\sin^{-1} x (1 + \frac{x^2}{\sqrt{1-x^2}})}{1-x^2} = \frac{\sin^{-1} x (\frac{\sqrt{1-x^2} + x^2}{\sqrt{1-x^2}})}{1-x^2}$.Wait,simplifying the expression: $\frac{dy}{dx} = \frac{\sin^{-1} x (1-x^2 + x^2)}{\sqrt{1-x^2}(1-x^2)} = \frac{\sin^{-1} x}{(1-x^2)^{3/2}}$.

If $y=\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}+\log \sqrt{1-x^2}$,then $\frac{d y}{d x}=$

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