Answer
$$\frac{{dy}}{{dx}} = - \frac{{2{{\operatorname{csch} }^{ - 1}}x}}{{\left| x \right|\sqrt {1 + {x^2}} }}$$
Work Step by Step
$$\eqalign{
& y = {\left( {{{\operatorname{csch} }^{ - 1}}x} \right)^2} \cr
& {\text{Differentiate both sides}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\left( {{{\operatorname{csch} }^{ - 1}}x} \right)}^2}} \right] \cr
& {\text{By the chain rule}} \cr
& \frac{{dy}}{{dx}} = 2\left( {{{\operatorname{csch} }^{ - 1}}x} \right)\frac{d}{{dx}}\left[ {{{\operatorname{csch} }^{ - 1}}x} \right] \cr
& {\text{Use Theorem 5}}{\text{.20 }}\frac{d}{{dx}}\left[ {{{\operatorname{csch} }^{ - 1}}x} \right] = - \frac{1}{{\left| x \right|\sqrt {1 + {x^2}} }} \cr
& \frac{{dy}}{{dx}} = 2\left( {{{\operatorname{csch} }^{ - 1}}x} \right)\left( { - \frac{1}{{\left| x \right|\sqrt {1 + {x^2}} }}} \right) \cr
& \frac{{dy}}{{dx}} = - \frac{{2{{\operatorname{csch} }^{ - 1}}x}}{{\left| x \right|\sqrt {1 + {x^2}} }} \cr} $$
