Answer
$\frac{{dy}}{{dx}} = \frac{{2x}}{{\left( {\arcsin {x^2}} \right)\sqrt {1 - {x^4}} }}$
Work Step by Step
$$\eqalign{
& y = \ln \left( {\arcsin {x^2}} \right) \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\ln \left( {\arcsin {x^2}} \right)} \right] \cr
& {\text{Use }}\frac{d}{{d\theta }}\left[ {\ln u} \right] = \frac{1}{u}\frac{{du}}{{dx}},{\text{ let }}u = \arcsin {x^2}{\text{ }} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\arcsin {x^2}}}\frac{d}{{dx}}\left[ {\arcsin {x^2}} \right] \cr
& {\text{Use }}\frac{d}{{d\theta }}\left[ {\arcsin u} \right] = \frac{1}{{\sqrt {1 - {u^2}} }}\frac{{du}}{{dx}},{\text{ let }}u = {x^2} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\arcsin {x^2}}}\left( {\frac{1}{{\sqrt {1 - {{\left( {{x^2}} \right)}^2}} }}} \right)\frac{d}{{dx}}\left[ {{x^2}} \right] \cr
& {\text{Compute the derivative and simplify}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\arcsin {x^2}}}\left( {\frac{1}{{\sqrt {1 - {x^4}} }}} \right)\left( {2x} \right) \cr
& \frac{{dy}}{{dx}} = \frac{{2x}}{{\left( {\arcsin {x^2}} \right)\sqrt {1 - {x^4}} }} \cr} $$
