The type of argument you describe is a big motivation for the definition of a continuous function. A basic result which captures the spirit of what you want is this: a continuous function $f : X \to Y$ between topological spaces is determined by its values on a dense subset of $X$. In particular, a continuous function $f : \mathbb{R} \to Y$ is determined by its values on the rationals. This gives rise to the following intuition:
Any "sufficiently continuous" statement you can prove for the rationals should be true of all real numbers.
The above result doesn't directly apply to this situation, at least not in a straightforward way: for one thing, knowing that the function $x^q$ exists for $q$ rational is not enough to immediately conclude that a function called $x^r$ even exists, let alone that it is continuous or differentiable. But it is still an important motivation for suspecting that such a thing ought to be true anyway.
For $x \ge 0$ it is a standard exercise in real analysis to prove that we can define
$$x^r = \lim_{n \to \infty} x^{q_n}$$
where $q_n$ is any sequence of rational numbers tending to $r$. Furthermore this function satisfies all of the expected algebraic identities, and so forth. But it is still not completely trivial to see that $x^r$ is differentiable for arbitrary real $r$ until you prove the basic properties of logarithms and exponentials, after which you then know that
$$x^r = e^{r \log x}.$$
Once you know this, you're in the money: the composite of differentiable functions is differentiable, and everything works out exactly the way you expect it to by the chain rule.
One might be tempted to argue as follows: surely if the function $x^r$ is a limit of a sequence of functions $x^{q_n}$, then the derivative of the function $x^r$ is a limit of the sequence of derivatives $q_n x^{q_n-1}$. But this argument is not so straightforward to formalize. First, there are various notions of convergence of a sequence of functions: there is pointwise convergence but also convergence with respect to various norms, and one must decide what notion is sensible here. Second, even if one takes a fairly restrictive definition of convergence (say uniform convergence), it is still false in general that a uniform limit of differentiable functions is differentiable (see Weierstrass function).
One needs the additional constraint that the derivatives themselves also converge uniformly. Now, the derivatives $q_n x^{q_n-1}$ do not converge uniformly to $rx^{r-1}$ on $\mathbb{R}_{\ge 0}$ (let's provisionally assume we only care about $r > 1$), but they do converge to $rx^{r-1}$ uniformly on the intervals $\left[ \frac{1}{k}, k \right]$ for all $k \in \mathbb{N}$ (say). I think that is enough to get this result without going through the stuff above about logarithms and exponentials, but I haven't checked it closely.
