Answer
$y^{\prime}=$ csch $x$
Work Step by Step
Apply the Chain Rule and
Th. 5.18:$\displaystyle \qquad \frac{d}{dx}[\tanh u]=({\rm sech}^{2}u)u^{\prime}$
$y=u(v(x))=\displaystyle \ln(\tanh\frac{x}{2})$
$y^{\prime}=\displaystyle \frac{d}{dx}[u(v(x))]=\frac{du}{dv}\cdot\frac{dv}{dx}$
$=[\displaystyle \frac{1}{\tanh(x/2)}][{\rm sech}^{2}(\frac{x}{2})\cdot\frac{1}{2}]$
$=\displaystyle \frac{1}{2\sinh(\frac{x}{2})\cosh(\frac{x}{2})}$
... apply the identity:$\quad\sinh 2x=2\sinh x\cosh x$
$=\displaystyle \frac{1}{\sinh(2\cdot\frac{x}{2})}$
$=\displaystyle \frac{1}{\sinh(x)}$
$=$ csch $x$
