Lösung 1.2:1b
Aus Online Mathematik Brückenkurs 2
We use the product rule with \displaystyle x^2 and \displaystyle \ln x as factors,
| \displaystyle \begin{align}
\bigl(x^{2}\ln x\bigr)' &= \bigl(x^2\bigr)'\,\ln x + x^2\,(\ln x)'\\[5pt] &= 2x\cdot\ln x + x^2\cdot\frac{1}{x}\\[5pt] &= 2x\ln x + x\,\textrm{.} \end{align} |
