Answer
$y' = \frac{{2\left( {{{\tan }^{ - 1}}x} \right)}}{{1 + {x^2}}}$
Work Step by Step
$$\eqalign{
& y = {\left( {{{\tan }^{ - 1}}x} \right)^2} \cr
& {\text{Differentiate}} \cr
& y' = \frac{d}{{dx}}\left[ {{{\left( {{{\tan }^{ - 1}}x} \right)}^2}} \right] \cr
& {\text{Use the general power rule for derivatives}} \cr
& y' = 2\left( {{{\tan }^{ - 1}}x} \right)\frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}x} \right] \cr
& {\text{Use the Derivatives of Inverse Trigonometric Functions }} \cr
& y' = 2\left( {{{\tan }^{ - 1}}x} \right)\left( {\frac{1}{{1 + {x^2}}}} \right) \cr
& {\text{Simplify}} \cr
& y' = \frac{{2\left( {{{\tan }^{ - 1}}x} \right)}}{{1 + {x^2}}} \cr} $$
