Lösung 1.1:2c
Aus Online Mathematik Brückenkurs 2
(Unterschied zwischen Versionen)
K |
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) |
||
| Zeile 1: | Zeile 1: | ||
We differentiate term by term, | We differentiate term by term, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
f^{\,\prime}(x) &= \frac{d}{dx}\,\bigl(e^x-\ln x\bigr)\\[5pt] | f^{\,\prime}(x) &= \frac{d}{dx}\,\bigl(e^x-\ln x\bigr)\\[5pt] | ||
&= \frac{d}{dx}\,e^{x} - \frac{d}{dx}\,\ln x\\[5pt] | &= \frac{d}{dx}\,e^{x} - \frac{d}{dx}\,\ln x\\[5pt] | ||
Version vom 12:50, 10. Mär. 2009
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.
