I had been reading about Lanzcos derivatives. So if $X = 0,1$ is a Bernoulli trial with each outcome equally likely,

$$ \frac{\mathrm{Cov}(f(t+hX), hX)}{\sigma_{hX}^2} = \frac{f(t+h)-f(t)}{h}$$

and the Lanzcos derivative is good on empirically defined functions. If $X \in [-1,1]$ uniformly at random,

$$ \frac{\mathrm{Cov}(f(t+hX), hX)}{\sigma_{hX}^2} = \frac{3}{2h^3}\int_{-1}^1 f(t+h)t \, dh.$$

Here differentiation becomes a linear regression problem. The derivative is defined as the Covariance of $f(t + hX)$ with a line $hX$ normalized by the variance.

Maybe you can generalize to things other than lines. Could there be a sense in which $$f(k) = \displaystyle \sum_{k=1}^n \phi(k) = \frac{1}{2}\left(1 + \sum_{k=1}^n \mu(k) \left\lfloor \frac{n}{k}\right\rfloor^2 \right)$$ can be differentiated with respect to Mobius function $\mu(\cdot)$ to get just $$\frac{df}{d\mu} =_? \displaystyle \frac{1}{2} \sum_{k=1}^n \left\lfloor \frac{n}{k}\right\rfloor^2 $$ I imagine some far-fetched sense of derivative should exists since $\phi(\cdot)$ and $\mu(\cdot)$ behave oddly at similar points.

I have evaded the issue of what it means for a function $f: \mathbb{Z} \to \mathbb{Z}$ to be ``differentiable". In the spirit of Coarse Geometry maybe we could define $F(x) = \frac{1}{n^2} f(\lfloor n x \rfloor)$. This question is a disaster but I thought I'd ask anyway if this or something similar could be made rigorous.