Find d/dx[x^x]

[hank]

Wowzahs...

I was thinking that this would be x * x^(x - 1), but I don't think I'm correct.

Is that right? Feels like there should be an e in there somewhere..

 

[pka]

First of all, you must know that for $\displaystyle x^x$ is defined only if $\displaystyle x > 0.$
That restriction means that you can use logarithms.
$\displaystyle f(x) = x^x \quad \Rightarrow \quad f'(x) = \left( {x^x } \right)\left[ {\ln (x) + 1} \right].$