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Lösung 1.2:3f

Aus Online Mathematik Brückenkurs 2

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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
-
{{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

ab=elnab=eblna

which, in our case, gives

xtanx=etanxlnx. (*)

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

ddxetanxlnx=etanxlnxtanxlnx 

and then the product rule

=etanxlnx(tanx)lnx+tanx(lnx)=etanxlnx1cos2xlnx+tanxx1=etanxlnxlnxcos2x+xtanx=xtanxlnxcos2x+xtanx

where we have used (*) in reverse.

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