Answer
$y' = \frac{{2x}}{{1 + {x^4}}}$
Work Step by Step
$$\eqalign{
& y = {\tan ^{ - 1}}\left( {{x^2}} \right) \cr
& {\text{Differentiate}} \cr
& y' = \frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\left( {{x^2}} \right)} \right] \cr
& {\text{Use the Derivatives of Inverse Trigonometric Functions }} \cr
& {\text{and the chain rule}}{\text{. }}\frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}u} \right] = \frac{1}{{{u^2} + 1}}\frac{{du}}{{dx}},\;{\text{ }}u = {x^2} \cr
& y' = \frac{1}{{1 + {{\left( {{x^2}} \right)}^2}}}\frac{d}{{dx}}\left[ {{x^2}} \right] \cr
& {\text{Therefore}}{\text{,}} \cr
& y' = \frac{1}{{1 + {x^4}}}\left( {2x} \right) \cr
& {\text{Simplify}} \cr
& y' = \frac{{2x}}{{1 + {x^4}}} \cr} $$
