Lösung 1.2:2d

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 1: Zeile 1:
We can see the expression as "ln of something",
We can see the expression as "ln of something",
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{{Displayed math||<math>\ln \bbox[#FFEEAA;,1.5pt]{\phantom{\ln x}}\,,</math>}}
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{{Abgesetzte Formel||<math>\ln \bbox[#FFEEAA;,1.5pt]{\phantom{\ln x}}\,,</math>}}
where "something" is <math>\ln x</math>.
where "something" is <math>\ln x</math>.
Zeile 7: Zeile 7:
Because we have a compound expression, we use the chain rule and obtain, roughly speaking, the outer derivative multiplied by the inner derivative,
Because we have a compound expression, we use the chain rule and obtain, roughly speaking, the outer derivative multiplied by the inner derivative,
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{{Displayed math||<math>\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)'\,,</math>}}
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{{Abgesetzte Formel||<math>\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)'\,,</math>}}
where the first factor on the right-hand side <math>1/\bbox[#FFEEAA;,1.5pt]{\,\ln x\,}</math> is the outer derivative of <math>\ln \bbox[#FFEEAA;,1.5pt]{\,\ln x\,}</math> and the other factor <math>\bigl( \bbox[#FFEEAA;,1.5pt]{\,\ln x\,} \bigr)'</math> is the inner derivative. Thus, we get
where the first factor on the right-hand side <math>1/\bbox[#FFEEAA;,1.5pt]{\,\ln x\,}</math> is the outer derivative of <math>\ln \bbox[#FFEEAA;,1.5pt]{\,\ln x\,}</math> and the other factor <math>\bigl( \bbox[#FFEEAA;,1.5pt]{\,\ln x\,} \bigr)'</math> is the inner derivative. Thus, we get
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{{Displayed math||<math>\frac{d}{dx}\,\ln\ln x = \frac{1}{\ln x}\cdot \frac{1}{x} = \frac{1}{x\ln x}\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\frac{d}{dx}\,\ln\ln x = \frac{1}{\ln x}\cdot \frac{1}{x} = \frac{1}{x\ln x}\,\textrm{.}</math>}}

Version vom 12:53, 10. 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{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:2d“