Derivative of Euler Phi Function w/ resp to Mobius function

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.

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Shouldn't it be $\frac{df}{d\mu} = \displaystyle \frac{1}{2} \sum_{k=1}^n \left\lfloor \frac{n}{k}\right\rfloor^2$? – Aeryk Dec 29 '11 at 13:40
thanks. guess i'm trying to say, for differentiable function the "generic" local behavior is a line. I wonder if these number theory functions also have their own generic behavior on $\mathbb{Z}$. Maybe I'm just looking for p-adic continuity or something. – john mangual Dec 29 '11 at 17:12