Lösung 1.2:1b
Aus Online Mathematik Brückenkurs 2
(Unterschied zwischen Versionen)
K (Lösning 1.2:1b moved to Solution 1.2:1b: Robot: moved page) |
Version vom 14:22, 16. Sep. 2008
We use the product rule with \displaystyle x^{2} and \displaystyle \ln x as factors:
\displaystyle \begin{align}
& \left( x^{2}\ln x \right)^{\prime }=\left( x^{2} \right)^{\prime }\ln x+x^{2}\left( \ln x \right)^{\prime } \\
& \\
& =2x\centerdot \ln x+x^{2}\centerdot \frac{1}{x}=2x\ln x+x \\
\end{align}
