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Lösung 1.2:3f

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
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Zeile 8: Zeile 8:
<math>x^{\tan x}=e^{\tan x\centerdot \ln x}</math>
<math>x^{\tan x}=e^{\tan x\centerdot \ln x}</math>
-
 
+
(*)
Now, we obtain the derivative by first using the chain rule
Now, we obtain the derivative by first using the chain rule
Zeile 25: Zeile 25:
& =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>
 +
 +
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


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:


=etanxlnxtanxlnx+tanxlnx=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“