Points of differentiability of $f(x) = \sum\limits_{n : q_n < x} c_n$

Let, $\{q_n\}_{n \in \mathbb{N}}$ be an enumeration of rational numbers. Consider the function $f : \mathbb{R} \to \mathbb{R}$ given by, $$\displaystyle f(x) = \sum\limits_{n : q_n < x} c_n$$

where, $\displaystyle \sum\limits_{n=1}^{\infty} c_n$ is an absolutely convergent positive series. The function is clearly monotone increasing, discontinuous at rationals (with jump exactly $c_n$ at $x = q_n$) and continuous at irrationals.

$1.$ I wish to ask about the points of differentiability of $f$?

Since $f$ is monotone it should be differentiable a.e. but how do we identify these points of differentiability? (as in a way of representing this set in a compact way)

Intuitively, it seems they should be related to the particular enumeration of the rationals $\{q_n\}$ at hand. For example if we have an enumeration such that for $\alpha \in \mathbb{R \setminus Q}$, we have $q_n \notin (\alpha - \delta_N , \alpha + \delta_N)$ for $1 \le n \le N$ (i.e., say $|q_n - \alpha| > \delta_n$ for $n \in \mathbb{N}$ where, $\delta_n \downarrow 0^{+}$ as $n \to \infty$) and now if we impose further the 'nice' property:

$\displaystyle \frac{f(\alpha + \delta_N) - f(\alpha)}{\delta_N} = \frac{1}{\delta_N}\sum\limits_{n : q_n \in (\alpha, \alpha + \delta_N)} c_n \to \lambda$, (as $N \to \infty$) and similarly one for the left derivative, we have $f'(\alpha) = \lambda$.

So, intuitively I can see how to choose an enumeration that makes the derivative equal $\lambda$ at $x = \alpha$ (or blows up at $\alpha$, i.e., $\lambda = + \infty$).

To clarify what I am asking: Given an enumeration of rationals, how do we come up with relevant definitions/concepts relating to said enumeration, which helps us identify which $\alpha$'s we should expect to be a point of differentiablity.

$2.$ Is there a way to estimate the derivative at these points?

Has these questions been addressed/answered in literature before? I'd love it if I could get some reference in this matter. Thanks! :-)

Maybe my answer is not really in the direction you wanted but here it is anyway.

Your function is increasing so it has bounded variation and hence it's derivative (in the distributional sense) is a radon measure. In fact since your function is right continuous everywhere it's a Cumulative distribution function (after eventual renormalisation). Its derivative $\mu$ can be classified in 3 parts : $$\mu=\mu_{ac}+\mu_{d}+\mu_{w}$$ where the first term is absolutely continuous (with respect to lebesgue measure) the second term is discrete and the last term is weird (but more often called "singular without atoms" in the literature). In your case we just have $\mu=\mu_d=\sum c_n\delta_{q_n}$.

Let's go back to the usual differentiation : $\;\displaystyle \frac{f(x+h)-f(x)}{h}=\frac{1}{h}\mu(]x;x+h])=\frac{\mu(]x;x+h])}{\lambda(]x;x+h])}$.

Now we can use the theory of differentiation of measures to deal with your question. Using theorem 7.13 or 7.14 of Real and complex analysis by W. Rudin (3rd edition) we see that the derivative of $f$ is equal to the derivative of $\mu$ with respect to the Lebesgue measure and since $\mu$ is discrete we get $f'=0$ a.e. (with respect to the Lebesgue measure).