Aus Online Mathematik Brückenkurs 2

We differentiate term by term,

\displaystyle \begin{align} f^{\,\prime}(x) &= \frac{d}{dx}\,\bigl(e^x-\ln x\bigr)\\[5pt] &= \frac{d}{dx}\,e^{x} - \frac{d}{dx}\,\ln x\\[5pt] &= e^{x}-\frac{1}{x}\,\textrm{.} \end{align}

Note: Because \displaystyle \ln x is not defined for \displaystyle x\le 0 we assume implicitly that \displaystyle x > 0.