Answer
$h^{\prime}(x)=\sinh^{2}x$
Work Step by Step
Apply
Th. 5.18:$\displaystyle \qquad \frac{d}{dx}[\sinh u]=(\cosh u)u^{\prime}$
$h(x)=\displaystyle \frac{1}{4}\sinh 2x-\frac{x}{2}$
$h^{\prime}(x)=\displaystyle \cosh(2x)\cdot\frac{1}{2}-\frac{1}{2}$
$=\displaystyle \frac{\cosh(2x)-1}{2}$
...Apply the identity$\displaystyle \qquad\sinh^{2}x=\frac{-1+\cosh 2x}{2}$
$h^{\prime}(x)=\sinh^{2}x$
