Lösung 1.1:2c
Aus Online Mathematik Brückenkurs 2
(Unterschied zwischen Versionen)
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| - | We differentiate term by term | + | We differentiate term by term, |
| + | {{Displayed math||<math>\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}</math>}} | ||
| - | <math>\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}</math> | ||
| - | + | Note: Because <math>\ln x</math> is not defined for <math>x\le 0</math> we assume implicitly that <math>x > 0</math>. | |
| - | + | ||
| - | <math>\ | + | |
| - | is not defined for | + | |
| - | <math>x\le 0</math> | + | |
| - | we assume implicitly that | + | |
| - | <math>x>0</math>. | + | |
Version vom 11:52, 14. Okt. 2008
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.
