Answer
$$
y^{\prime}=\frac{d}{dx} \left[ y \right] = \frac{d}{dx} \left[ x \sin ^{-1} x+\sqrt{1-x^{2}} \right] =\sin ^{-1} x
$$
Work Step by Step
$$
y=x \sin ^{-1} x+\sqrt{1-x^{2}}
$$
Differentiating both sides of this equation we have
$$
\begin{aligned}
\frac{d}{dx} \left[ y \right] &= \frac{d}{dx} \left[ x \sin ^{-1} x+\sqrt{1-x^{2}} \right] \\
y^{\prime}&=x \cdot \frac{1}{\sqrt{1-x^{2}}}+\left(\sin ^{-1} x\right)(1)+\frac{1}{2}\left(1-x^{2}\right)^{-1 / 2}(-2 x) \\
&=\frac{x}{\sqrt{1-x^{2}}}+\sin ^{-1} x-\frac{x}{\sqrt{1-x^{2}}} \\
&=\sin ^{-1} x
\end{aligned}
$$
