Lösung 1.1:2c
Aus Online Mathematik Brückenkurs 2
We differentiate term by term:
\displaystyle \begin{align}
& {f}'\left( x \right)=\frac{d}{dx}\left( e^{x}-\ln x \right) \\
& =\frac{d}{dx}e^{x}-\frac{d}{dx}\ln x=e^{x}-\frac{1}{x} \\
\end{align}
NOTE: because
\displaystyle \text{ln }x
is not defined for
\displaystyle x\le 0
we assume implicitly that
\displaystyle x>0.
