Answer
$y'=sin^{-1}x$
Work Step by Step
$y=xsin^{-1}x+\sqrt{1-x^2}\\
y'=x\times{\frac{1}{\sqrt{1-x^2}}}+sin^{-1}x\times1+({-\frac{2x}{2\sqrt{1-x^2}}})\\
y'={\frac{x}{\sqrt{1-x^2}}}+sin^{-1}-\frac{x}{\sqrt{1-x^2}}\\
y'=sin^{-1}x$
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