How to interpret $x^{\alpha}$, for $\alpha \in (0, \infty)\diagdown \mathbb N$? Let $f:(0, \infty)\to \mathbb R$ such that $f(x)= x^{n}, \ (x\in \mathbb R , n\in \mathbb N$).(here we interpret $x^{n}$ as product of $x$ with itself $n$ times, that is, $x^{n}= x\cdot \cdot \cdot x$) Clearly, $f$ is differentiable and its derivative is given by,  $f'(x)= nx^{n-1}.$

My Questions: (I) Let $x\in (0, \infty)$. How to interpret $x^{\alpha}$, for $\alpha \in (0, \infty)\diagdown \mathbb N$ ? (II) Let $g:(0, \infty) \to \mathbb R$ such that $g(x)= x^{\alpha}, \ (\alpha \in (0, \infty)\diagdown \mathbb N).$ Can we expect $g$ is differentiable on $(0, \infty)$, and its derivative is given by $g'(x)= \alpha x^{\alpha -1}$ ?

Thanks,

The typical way is to define
$$ x^a = \exp(a \cdot \log{x}) $$
Since $\exp$ can be defined for very general spaces, and $\log$ as its inverse, this approach is very general (if perhaps unsatisfying).
To fully answer your question, you can use chain rule to recover your expected derivatives.
