Lösung.

Es wird

$ \mbox{$\displaystyle \begin{array}{l} (x^\alpha)' \;=\; (e^{\alpha\log x})' \... ... x)^2 \;=\; (\cosh x)^{-2} \;=\; 1-(\tanh x)^2\;.\vspace{1mm}\\ \end{array}$}$