Answer
$$y' = \frac{{2{{\coth }^{ - 1}}x}}{{1 - {x^2}}}$$
Work Step by Step
$$\eqalign{
& y = {\left( {{{\coth }^{ - 1}}x} \right)^2} \cr
& {\text{find the derivatvive by using the chain rule}} \cr
& y' = 2\left( {{{\coth }^{ - 1}}x} \right)\left( {{{\coth }^{ - 1}}x} \right)' \cr
& {\text{use the theorem 6}}{\text{.8}}{\text{.5 }}\left( {{\text{ see page 479}}} \right){\text{ }} \cr
& \frac{d}{{dx}}\left[ {{{\coth }^{ - 1}}u} \right] = \frac{1}{{1 - {u^2}}}\frac{{du}}{{dx}} \cr
& y' = 2\left( {{{\coth }^{ - 1}}x} \right)\left( {\frac{1}{{1 - {x^2}}}} \right) \cr
& {\text{simplifying}} \cr
& y' = \frac{{2{{\coth }^{ - 1}}x}}{{1 - {x^2}}} \cr} $$
