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Lösung 1.1:2c

Aus Online Mathematik Brückenkurs 2

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We differentiate term by term,
We differentiate term by term,
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{{Displayed math||<math>\begin{align}
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{{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,

f(x)=ddxex−lnx=ddxex−ddxlnx=ex−x1.


Note: Because lnx is not defined for x0 we assume implicitly that x0.

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.1:2c“