Mathematical proof of relationship between derivative and convolution
There's no mathematical proof per se; the fact is that finite differences are one method of approximating derivatives. Remember the definition of a derivative:

$$
f'(x) = \lim_{\Delta x \to 0} \frac{f(x+\Delta x) - f(x)}{\Delta x}
$$

The above equation states that to calculate the derivative of a function $f$ at position $x$, one takes the difference between the function evaluated at $x$ and the function evaluated at some small distance $\Delta x$ away. The ratio of the difference in the function value between the two points and the distance traveled along the $x$-axis is defined to be the derivative, as the distance between the two points becomes infinitesimally small.

A finite-difference derivative method, then, is just a way of approximating the above relationship using some finite spacing instead of the infinitely-small spacing implied by the limit. This type of approach makes sense for real-world applications where you might not analytically know the function $f(x)$  (and therefore can't evaluate its behavior across arbitrarily-small intervals.

With that said, there are a number of ways to approximate the derivative of a signal in this way; one example is the causal first-order difference:

$$
x'(t) \approx \frac{x(t) - x(t-T)}{T}
$$

This is simple to implement, but performs best if $T$ is small relative to the inverse of the signal's bandwidth. A slightly more complicated method can provide better results in some cases (the second-order difference):

$$
x'(t) \approx \frac{x(t) - 2x(t-T) +x(t-2T)}{T^2}
$$

I've used the "backward second-order difference" form above to show how they could be applied using causal filters, but note that they will have a resulting delay commensurate with the order of the approximation.