the derivative of the function defined on $\mathbb{R}$ by $f (x) = x^z$
I want to show that the derivative of the function defined on $\mathbb{R}$ by $f (x) = x^z$ with $z\in \mathbb{C}$ is $zx^{z-1}$. For that I tried to apply the definition, let $h\neq 0$ then $$ \frac{f(x+h)-f(x)}{h}=\frac{(x+h)^z-x^z}{h} $$ but how can I develop the expression $(x+h)^z$ to get the result?
If you like, the calculation can be reduced to the derivative of the exponential function as follows:
$\log x^z = z\log x$, so $x^z = \exp(z\log x)$. Applying the chain rule gives the derivative as $$ \frac d{dx} x^z = \frac zx \exp(z\log x) = zx^{z-1}. $$.