Processing Math: Done
Lösung 1.2:3f
Aus Online Mathematik Brückenkurs 2
(Unterschied zwischen Versionen)
K (Lösning 1.2:3f moved to Solution 1.2:3f: Robot: moved page) |
|||
Zeile 1: | Zeile 1: | ||
- | {{ | + | We have no differentiation rule for a function raised to another function, but instead we rewrite |
- | < | + | |
- | {{ | + | |
+ | <math>a^{b}=e^{\ln a^{b}}=e^{b\ln a}</math>, | ||
+ | |||
+ | which, in our case, gives | ||
+ | |||
+ | |||
+ | <math>x^{\tan x}=e^{\tan x\centerdot \ln x}</math> | ||
+ | |||
+ | |||
+ | Now, we obtain the derivative by first using the chain rule | ||
+ | |||
+ | |||
+ | <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> | ||
+ | |||
+ | |||
+ | and then the product rule: | ||
+ | |||
+ | |||
+ | <math>\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}</math> |
Version vom 13:41, 12. Okt. 2008
We have no differentiation rule for a function raised to another function, but instead we rewrite
which, in our case, gives
lnx
Now, we obtain the derivative by first using the chain rule
tanx
lnx
=e
tanx
lnx
tanx
lnx
and then the product rule:
lnx
tanx
lnx+tanx
lnx
=etanx
lnx
1cos2x
lnx+tanx
x1
=etanx
lnx
lnxcos2x+xtanx
=xtanx
lnxcos2x+xtanx