Lösung 1.2:3f

Aus Online Mathematik Brückenkurs 2

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<math>x^{\tan x}=e^{\tan x\centerdot \ln x}</math>
<math>x^{\tan x}=e^{\tan x\centerdot \ln x}</math>
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(*)
Now, we obtain the derivative by first using the chain rule
Now, we obtain the derivative by first using the chain rule
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& =x^{\tan x}\left( \frac{\ln x}{\cos ^{2}x}+\frac{\tan x}{x} \right) \\
& =x^{\tan x}\left( \frac{\ln x}{\cos ^{2}x}+\frac{\tan x}{x} \right) \\
\end{align}</math>
\end{align}</math>
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where we have used (*) in reverse.

Version vom 14:55, 13. Okt. 2008

We have no differentiation rule for a function raised to another function, but instead we rewrite


\displaystyle a^{b}=e^{\ln a^{b}}=e^{b\ln a},

which, in our case, gives


\displaystyle x^{\tan x}=e^{\tan x\centerdot \ln x} (*)

Now, we obtain the derivative by first using the chain rule


\displaystyle \frac{d}{dx}e^{\left\{ \left. \tan x\centerdot \ln x \right\} \right.}=e^{\left\{ \left. \tan x\centerdot \ln x \right\} \right.}\centerdot \left( \left\{ \left. \tan x\centerdot \ln x \right\} \right. \right)^{\prime }


and then the product rule:


\displaystyle \begin{align} & =e^{\tan x\centerdot \ln x}\left( \left( \tan x \right)^{\prime }\centerdot \ln x+\tan x\centerdot \left( \ln x \right)^{\prime } \right) \\ & =e^{\tan x\centerdot \ln x}\left( \frac{1}{\cos ^{2}x}\centerdot \ln x+\tan x\centerdot \frac{1}{x} \right) \\ & =e^{\tan x\centerdot \ln x}\left( \frac{\ln x}{\cos ^{2}x}+\frac{\tan x}{x} \right) \\ & =x^{\tan x}\left( \frac{\ln x}{\cos ^{2}x}+\frac{\tan x}{x} \right) \\ \end{align}

where we have used (*) in reverse.

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_1.2:3f“