Answer
$$\frac{{dy}}{{dx}} = \frac{{{{\sec }^2}\left( {\arcsin x} \right)}}{{\sqrt {1 - {x^2}} }}$$
Work Step by Step
$$\eqalign{
& y = \tan \left( {\arcsin x} \right) \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\tan \left( {\arcsin x} \right)} \right] \cr
& {\text{By the chain rule}} \cr
& \frac{{dy}}{{dx}} = {\sec ^2}\left( {\arcsin x} \right)\frac{d}{{dx}}\left[ {\arcsin x} \right] \cr
& \frac{{dy}}{{dx}} = {\sec ^2}\left( {\arcsin x} \right)\left( {\frac{1}{{\sqrt {1 - {x^2}} }}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{{{{\sec }^2}\left( {\arcsin x} \right)}}{{\sqrt {1 - {x^2}} }} \cr} $$
