Answer
$$\frac{{dy}}{{dx}} = \frac{{2x}}{{1 - {x^4}}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {\coth ^{ - 1}}\left( {{x^2}} \right) \cr
& {\text{Differentiate both sides}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\coth }^{ - 1}}\left( {{x^2}} \right)} \right] \cr
& {\text{Use Theorem 5}}{\text{.20 }}\frac{d}{{dx}}\left[ {{{\coth }^{ - 1}}u} \right] = \frac{{u'}}{{1 - {u^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{\frac{d}{{dx}}\left[ {{x^2}} \right]}}{{1 - {{\left( {{x^2}} \right)}^2}}} \cr
& {\text{Compute derivatives}} \cr
& \frac{{dy}}{{dx}} = \frac{{2x}}{{1 - {x^4}}} \cr} $$
