Lösung 1.2:2d
Aus Online Mathematik Brückenkurs 2
(Unterschied zwischen Versionen)
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) |
K (Solution 1.2:2d moved to Lösung 1.2:2d: Robot: moved page) |
Version vom 10:04, 11. Mär. 2009
We can see the expression as "ln of something",
| \displaystyle \ln \bbox[#FFEEAA;,1.5pt]{\phantom{\ln x}}\,, |
where "something" is \displaystyle \ln x.
Because we have a compound expression, we use the chain rule and obtain, roughly speaking, the outer derivative multiplied by the inner derivative,
| \displaystyle \frac{d}{dx}\,\ln \bbox[#FFEEAA;,1.5pt]{\,\ln x\,} = \frac{1}{\bbox[#FFEEAA;,1.5pt]{\,\ln x\,}}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\,\ln x\,} \bigr)'\,, |
where the first factor on the right-hand side \displaystyle 1/\bbox[#FFEEAA;,1.5pt]{\,\ln x\,} is the outer derivative of \displaystyle \ln \bbox[#FFEEAA;,1.5pt]{\,\ln x\,} and the other factor \displaystyle \bigl( \bbox[#FFEEAA;,1.5pt]{\,\ln x\,} \bigr)' is the inner derivative. Thus, we get
| \displaystyle \frac{d}{dx}\,\ln\ln x = \frac{1}{\ln x}\cdot \frac{1}{x} = \frac{1}{x\ln x}\,\textrm{.} |
