Answer
$$\frac{{dy}}{{dx}} = {\sin ^{ - 1}}x $$
Work Step by Step
$$\eqalign{
& y = x{\sin ^{ - 1}}x + \sqrt {1 - {x^2}} \cr
& {\text{find the derivative of }}y{\text{ with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{{d\left( {x{{\sin }^{ - 1}}x} \right)}}{{dx}} + \frac{{d\left( {\sqrt {1 - {x^2}} } \right)}}{{dx}} \cr
& {\text{product rule}} \cr
& \frac{{dy}}{{dx}} = x\frac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right) + {\sin ^{ - 1}}x\frac{{d\left( x \right)}}{{dx}} + \frac{{d\left( {\sqrt {1 - {x^2}} } \right)}}{{dx}} \cr
& {\text{ use the formulas }} \cr
& \frac{{d\left( {{{\sin }^{ - 1}}u} \right)}}{{dx}} = \frac{1}{{\sqrt {1 - {u^2}} }}\frac{{du}}{{dx}},\left( {{\text{see table 7}}{\text{.3}}} \right)\cr
& {\text{ and the chain rule}} \cr
& \frac{{dy}}{{dx}} = x\left( {\frac{1}{{\sqrt {1 - {x^2}} }}} \right) + {\sin ^{ - 1}}x\left( 1 \right) + \left( {\frac{1}{2}} \right){\left( {1 - {x^2}} \right)^{ - 1/2}}\left( { - 2x} \right) \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = \frac{x}{{\sqrt {1 - {x^2}} }} + {\sin ^{ - 1}}x - x{\left( {1 - {x^2}} \right)^{ - 1/2}} \cr
& \frac{{dy}}{{dx}} = \frac{x}{{\sqrt {1 - {x^2}} }} + {\sin ^{ - 1}}x - \frac{x}{{\sqrt {1 - {x^2}} }} \cr
& \frac{{dy}}{{dx}} = {\sin ^{ - 1}}x \cr} $$
