I hope you can help me with this one because I seem to not quiet get a start here :/
Lets say we got a $b\in\mathbb{R}_{\gt 0}$ and a $y\in\mathbb{R}$ and we define $b^y:=\exp\left(\ln b \cdot y\right)$.
How do I prove that the derivation of a function $$f:\mathbb{R}_{\gt 0}\rightarrow\mathbb{R}, x\mapsto x^a$$ with a well defined $a\in\mathbb{R}$ is $f'(x) = ax^{a-1}$?
By definition, we have $x^a=\exp(a\ln x)$. Differentiate, using the Chain Rule. We get $$\frac{a}{x}\exp(a\ln x).\tag{1}$$ Using $\frac{1}{x}=\exp(-\ln x)$, we can rewrite (1) as $a\exp((a-1)\ln x)$, that is, as $ax^{a-1}$.