Lösung 1.2:2d

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We can see the expression as "ln of something",
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<center> [[Image:1_2_2d.gif]] </center>
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<math>\ln \left\{ \left. {} \right\} \right.</math>,
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where "something" is
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<math>\ln x</math>.
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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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<math>\frac{d}{dx}\ln \left\{ \left. \ln x \right\} \right.=\frac{1}{\ln x}\centerdot \left( \left\{ \left. \ln x \right\} \right. \right)^{\prime }</math>
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where the first factor on the right-hand side
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<math>\frac{1}{\ln x}</math>
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is the outer derivative of
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<math>\ln \left\{ \left. \ln x \right\} \right.</math>
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and the other factor
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<math>\left( \left\{ \left. \ln x \right\} \right. \right)^{\prime }</math>
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is the inner derivative. Thus, we get
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<math>\frac{d}{dx}\ln \left\{ \left. \ln x \right\} \right.=\frac{1}{\ln x}\centerdot \frac{1}{x}=\frac{1}{x\ln x}</math>

Version vom 13:13, 11. Okt. 2008

We can see the expression as "ln of something",


\displaystyle \ln \left\{ \left. {} \right\} \right.,

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 \left\{ \left. \ln x \right\} \right.=\frac{1}{\ln x}\centerdot \left( \left\{ \left. \ln x \right\} \right. \right)^{\prime }


where the first factor on the right-hand side \displaystyle \frac{1}{\ln x} is the outer derivative of \displaystyle \ln \left\{ \left. \ln x \right\} \right. and the other factor \displaystyle \left( \left\{ \left. \ln x \right\} \right. \right)^{\prime } is the inner derivative. Thus, we get


\displaystyle \frac{d}{dx}\ln \left\{ \left. \ln x \right\} \right.=\frac{1}{\ln x}\centerdot \frac{1}{x}=\frac{1}{x\ln x}

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