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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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The correct answer is:
$\frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}}$
We have, $y=\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}+\log \sqrt{1-x^2}$
$\left[1 \cdot \sin ^{-1} x+\frac{x}{\sqrt{1-x^2}}\right] \sqrt{1-x^2}-x \sin ^{-1} x$
$\begin{aligned} & \therefore \frac{d y}{d x}=\frac{\left.\left(\frac{1}{2 \sqrt{1-x^2}}(-2 x)\right)\right]}{\left(1-x^2\right)} \\ & +\frac{1}{2} \frac{1}{\left(1-x^2\right)}(-2 x) \\ & \end{aligned}$
$\begin{aligned} & =\frac{1}{1-x^2}\left[\sqrt{1-x^2} \sin ^{-1} x+x+\frac{x^2 \sin ^{-1} x}{\sqrt{1-x^2}}-x\right] \\ & =\frac{1}{1-x^2}\left[\frac{\left(1-x^2\right) \sin ^{-1} x+x^2 \sin ^{-1} x}{\sqrt{1-x^2}}\right]=\frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}}\end{aligned}$
$\left[1 \cdot \sin ^{-1} x+\frac{x}{\sqrt{1-x^2}}\right] \sqrt{1-x^2}-x \sin ^{-1} x$
$\begin{aligned} & \therefore \frac{d y}{d x}=\frac{\left.\left(\frac{1}{2 \sqrt{1-x^2}}(-2 x)\right)\right]}{\left(1-x^2\right)} \\ & +\frac{1}{2} \frac{1}{\left(1-x^2\right)}(-2 x) \\ & \end{aligned}$
$\begin{aligned} & =\frac{1}{1-x^2}\left[\sqrt{1-x^2} \sin ^{-1} x+x+\frac{x^2 \sin ^{-1} x}{\sqrt{1-x^2}}-x\right] \\ & =\frac{1}{1-x^2}\left[\frac{\left(1-x^2\right) \sin ^{-1} x+x^2 \sin ^{-1} x}{\sqrt{1-x^2}}\right]=\frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}}\end{aligned}$
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