Answer
$y' = {\sin ^{ - 1}}x$
Work Step by Step
$$\eqalign{
& y = x{\sin ^{ - 1}}x + \sqrt {1 - {x^2}} \cr
& {\text{Differentiate}} \cr
& y' = \frac{d}{{dx}}\left[ {x{{\sin }^{ - 1}}x + \sqrt {1 - {x^2}} } \right] \cr
& y' = \frac{d}{{dx}}\left[ {x{{\sin }^{ - 1}}x} \right] + \frac{d}{{dx}}\left[ {\sqrt {1 - {x^2}} } \right] \cr
& {\text{Use the product rule}} \cr
& y' = x\frac{d}{{dx}}\left[ {{{\sin }^{ - 1}}x} \right] + {\sin ^{ - 1}}x\frac{d}{{dx}}\left[ x \right] + \frac{d}{{dx}}\left[ {\sqrt {1 - {x^2}} } \right] \cr
& {\text{Compute deivatives}} \cr
& y' = x\left( {\frac{1}{{\sqrt {1 - {x^2}} }}} \right) + {\sin ^{ - 1}}x\left( 1 \right) + \frac{{ - 2x}}{{2\sqrt {1 - {x^2}} }} \cr
& {\text{Simplify}} \cr
& y' = {\sin ^{ - 1}}x \cr} $$
