Answer
$$\frac{{dy}}{{dx}} = - 16x{\operatorname{csch} ^2}\left( {8{x^2}} \right)$$
Work Step by Step
$$\eqalign{
& y = \coth \left( {8{x^2}} \right) \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\coth \left( {8{x^2}} \right)} \right] \cr
& {\text{Using the rules of theorem 5}}{\text{.18 }}\left( {{\text{Page 385}}} \right) \cr
& \frac{{dy}}{{dx}} = - {\operatorname{csch} ^2}\left( {8{x^2}} \right)\frac{d}{{dx}}\left[ {8{x^2}} \right] \cr
& \frac{{dy}}{{dx}} = - {\operatorname{csch} ^2}\left( {8{x^2}} \right)\left( {16x} \right) \cr
& \frac{{dy}}{{dx}} = - 16x{\operatorname{csch} ^2}\left( {8{x^2}} \right) \cr} $$
