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