Here's what i'm trying to prove:
Let $f: (0, +\infty) \to (0,+\infty)$ be our function defined:
$$\forall x \in \mathbb{R}^+: f(x) = x^r$$
where $r \in \mathbb{Q} \setminus \{0\}$. Prove that $f$ is differentiable and find its derivative.
Proof Attempt:
Let $r \in \mathbb{Q} \setminus \{0\}$. Then, $r = \frac{m}{n}$, where we will say that $n > 0$. Both $m$ and $n$ will be integers, obviously. We will also define $I = (0,+\infty)$ since it is mentioned alot.
Define two functions $h:I \to I$ and $g: I \to I$ as follows:
$$\forall x \in I: h(x) = x^m$$
$$\forall x \in I: g(x) = x^{\frac{1}{n}}$$
Now, both of these functions are bijective and we know that $f$, defined above, is bijective. We can also see that $f = h \circ g$. So, proving the differentiability of $f$ is tantamount to proving the differentiability of $h$ and $g$.
$h$ being differentiable in $I$ is a simple consequence of the quotient rule. Now, consider $g$. Clearly, $g^{-1}$ is differentiable in $I$.
Since $g^{-1}$ is continuous and increasing in $I$, it follows that $g$ is continuous and increasing in $I$. It is also the case that $g^{-1}$ has a nonzero derivative at every point in $I$. Hence, by the inverse function theorem, it follows that $g$ is differentiable in $I$.
Hence, $f$ is differentiable in $I$. Now, we use the chain rule and deduce that for every $x \in I$:
$$f'(x) = (h(g(x)))' = h'(g(x))g'(x)$$
All that's left is to compute $g'(x)$. Let $y = x^{\frac{1}{n}}$. Then, the derivative of $g(x)$ is given to us by the inverse function theorem:
$$g'(x) = \frac{1}{(g^{-1})'(y)} = \frac{1}{ny^{n-1}} = \frac{1}{n}\cdot y^{1-n} = \frac{1}{n} \cdot x^{\frac{1}{n}-1}$$
and obviously, $h'(g(x)) = m \cdot x^{\frac{m}{n}-\frac{1}{n}}$. So, combining these results, we have:
$$f'(x) = \frac{m}{n} \cdot x^{\frac{m}{n}-1} = rx^{r-1}$$
which was the desired result.
Does this proof work? If not, why? How can I fix it? Also, I used quite a few results that I've already proved so I hope that I've stated them clearly enough in my argument above.
Edit:
So, what I meant by $h$ being differentiable because of the quotient rule is the following;
If $m > 0$, then $x^m$ is differentiable by way of the regular power rule.
If $m = 0$, then $x^m = 1$ and that is certainly differentiable.
If $m < 0$, then let $n = -m$. Clearly, $x^m = x^{-n} = \frac{1}{x^n}$. Since $x^n$ is differentiable, by the power rule, and since $1$ is differentiable, it follows that their quotient is differentiable by the quotient rule. We also note that $x^n \neq 0$ in the interval $(0,+\infty)$ so the hypotheses of the rule are satisfied.