Aus Online Mathematik Brückenkurs 2

We have no differentiation rule for a function raised to another function, but instead we rewrite

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

which, in our case, gives

\displaystyle x^{\tan x}=e^{\tan x\centerdot \ln x} (*)

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

\displaystyle \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 }

and then the product rule:

\displaystyle \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}

where we have used (*) in reverse.