Chapter 8 - Techniques of Integration - 8.4 Integrals Involving Hyperbolic and Inverse Hyperbolic Functions - Exercises - Page 415: 45

$$\frac{d}{d y} g d(y) =\operatorname{sech} y$$

Since $$g d(y)=\tan ^{-1}(\sinh y)$$ Then \begin{aligned} \frac{d}{d y} g d(y) &=\frac{d}{d y} \tan ^{-1}(\sinh y) \\ &=\frac{1}{1+\sinh ^{2} y} \cosh y \\ &=\frac{1}{\cosh ^{2} y} \cosh y \\ &=\frac{1}{\cosh y}\\ &=\operatorname{sech} y \end{aligned}