Does the "average derivative of a function" exist? Given a function $f$ that is differentiable everywhere,
what would
$$\frac{\sum_{x\in\mathbb{R}}\frac{df(x)}{dx}}{|\mathbb{R}|}$$
denote? Does this expression even make sense? The size of $\mathbb{R}$ is infinite, so there is more confusion there as well.
For example, intuitively, I know
$$f(x)=x$$
has an average derivative of $1$, since $\frac{dx}{dx}=1$ for all $x\in\mathbb{R}$. How can this be extended to all differentiable functions?
 A: The average value of a function $g$, denoted $g_{\text{avg}}$, is typically defined via its integral:
$$g_{\text{avg over }[a,b]} = \frac{1}{b-a} \int_a^b g(x)\,dx.$$
So if you want the average value of $f'$ for some function $f$ over $[a,b]$, this would be given by
$$f'_{\text{avg over }[a,b]} = \frac{1}{b-a} \int_a^b f'(x)\,dx = \frac{f(b)-f(a)}{b-a}.$$
Note that this is just the difference quotient representing the slope of the straight line between the points $(a, f(a))$ and $(b, f(b))$. You could then take a limit as $a\to-\infty$ and $b\to\infty$ to get what you're looking for, so
$$f'_{\text{avg over }\mathbb{R}} = \lim_{(a,b)\to(-\infty,\infty)} \frac{f(b)-f(a)}{b-a}.$$
If you want to regularize a bit, you could drop to a single limit a la a Cauchy principal value:
$$PV\, f'_{\text{avg over }\mathbb{R}} = \lim_{a\to\infty} \frac{f(a)-f(-a)}{a - (-a)} = \lim_{a\to\infty} \frac{f(a)-f(-a)}{2a}.$$
Fair warning: a lot of functions will not have an average derivative over all of $\mathbb{R}$ or even have a finite regularized average.
