Lösung 1.2:3f

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We have no differentiation rule for a function raised to another function, but instead we rewrite
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We have no differentiation rule for a function raised to another function, but instead we use the formula
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{{Displayed math||<math>a^b = e^{\ln a^b} = e^{b\ln a}\,,</math>}}
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<math>a^{b}=e^{\ln a^{b}}=e^{b\ln a}</math>,
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which, in our case, gives
which, in our case, gives
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{{Displayed math||<math>x^{\tan x} = e^{\tan x\cdot\ln x}\,\textrm{.}</math>|(*)}}
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<math>x^{\tan x}=e^{\tan x\centerdot \ln x}</math>
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(*)
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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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{{Displayed math||<math>\frac{d}{dx}\,e^{\bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}} = {}\rlap{e^{\bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}\bigr)'}\phantom{e^{\tan x\cdot \ln x}\bigl((\tan x)'\cdot\ln x + \tan x\cdot (\ln x)'\bigr)}</math>}}
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<math>\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 }</math>
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and then the product rule
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and then the product rule:
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<math>\begin{align}
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{{Displayed math||<math>\begin{align}
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& =e^{\tan x\centerdot \ln x}\left( \left( \tan x \right)^{\prime }\centerdot \ln x+\tan x\centerdot \left( \ln x \right)^{\prime } \right) \\
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\phantom{\frac{d}{dx}\,e^{\bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}}}{}
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& =e^{\tan x\centerdot \ln x}\left( \frac{1}{\cos ^{2}x}\centerdot \ln x+\tan x\centerdot \frac{1}{x} \right) \\
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&= e^{\tan x\cdot \ln x}\bigl((\tan x)'\cdot\ln x + \tan x\cdot (\ln x)'\bigr)\\[5pt]
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& =e^{\tan x\centerdot \ln x}\left( \frac{\ln x}{\cos ^{2}x}+\frac{\tan x}{x} \right) \\
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&= e^{\tan x\cdot\ln x}\Bigl(\frac{1}{\cos^2\!x}\cdot\ln x + \tan x\cdot\frac{1}{x} \Bigr)\\[5pt]
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& =x^{\tan x}\left( \frac{\ln x}{\cos ^{2}x}+\frac{\tan x}{x} \right) \\
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&= e^{\tan x\cdot\ln x}\Bigl(\frac{\ln x}{\cos^2\!x} + \frac{\tan x}{x}\Bigr)\\[5pt]
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\end{align}</math>
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&= x^{\tan x}\Bigl(\frac{\ln x}{\cos^2\!x} + \frac{\tan x}{x}\Bigr)\,,
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\end{align}</math>}}
where we have used (*) in reverse.
where we have used (*) in reverse.

Version vom 13:24, 15. Okt. 2008

We have no differentiation rule for a function raised to another function, but instead we use the formula

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

which, in our case, gives

\displaystyle x^{\tan x} = e^{\tan x\cdot\ln x}\,\textrm{.} (*)

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

\displaystyle \frac{d}{dx}\,e^{\bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}} = {}\rlap{e^{\bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}}\cdot \bigl( \bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}\bigr)'}\phantom{e^{\tan x\cdot \ln x}\bigl((\tan x)'\cdot\ln x + \tan x\cdot (\ln x)'\bigr)}

and then the product rule

\displaystyle \begin{align}

\phantom{\frac{d}{dx}\,e^{\bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}}}{} &= e^{\tan x\cdot \ln x}\bigl((\tan x)'\cdot\ln x + \tan x\cdot (\ln x)'\bigr)\\[5pt] &= e^{\tan x\cdot\ln x}\Bigl(\frac{1}{\cos^2\!x}\cdot\ln x + \tan x\cdot\frac{1}{x} \Bigr)\\[5pt] &= e^{\tan x\cdot\ln x}\Bigl(\frac{\ln x}{\cos^2\!x} + \frac{\tan x}{x}\Bigr)\\[5pt] &= x^{\tan x}\Bigl(\frac{\ln x}{\cos^2\!x} + \frac{\tan x}{x}\Bigr)\,, \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“