Lösung 1.2:3f

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We have no differentiation rule for a function raised to another function, but instead we use the formula
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>}}
+
{{Abgesetzte Formel||<math>a^b = e^{\ln a^b} = e^{b\ln a}\,,</math>}}
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>|(*)}}
+
{{Abgesetzte Formel||<math>x^{\tan x} = e^{\tan x\cdot\ln x}\,\textrm{.}</math>|(*)}}
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>}}
+
{{Abgesetzte Formel||<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>}}
and then the product rule
and then the product rule
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\phantom{\frac{d}{dx}\,e^{\bbox[#FFEEAA;,1.5pt]{\tan x\cdot\ln x}}}{}
\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((\tan x)'\cdot\ln x + \tan x\cdot (\ln x)'\bigr)\\[5pt]

Version vom 12:54, 10. Mär. 2009

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.

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