Derivative of $f(x) = a \sum_{i=1}^x 10^{x-i}$ Consider the following function:
$f(x) = a \sum_{i=1}^x 10^{x-i}= a (10^{x-1} + 10^{x-2} + \cdots + 10^{x-x})$
whose domain are the positive integers greater than 1.
What is its derivative function?
Despite the function being non-continuous I think it should be possible to find its derivative, but I don't know how to do it analytically.
 A: First let me note that your question, as it stands, does not make sense. Assume we are looking for a function $g(x)$ such that:


*

*$g(x)$ is differentiable

*$g(x) = f(x)$  for all $x \in \mathbb N$.


This is not unique. Consider the following process to form another function $g^\ast (x)$, take such a $g(x)$ and for all integer $n$, consider the interval $[n + 0.4, n + 0.6]$, and just slightly change your sketch (like add a hump or something). Clearly, this new function $g^\ast(x) $ still satisfies your conditions. We can deform $g(x)$ in different ways and still maintain your properties.
If you are asking for something like $\Gamma(x)$ is to $(x-1)!$, then you still need conditions. There are many such functions which interlope the factorial function, the gamma function just has some useful properties that we like (it is log-convex). The Bohr-Mollerup theorem proves that the gamma function is the only such function satisfying these properties at the integers. That is why we use it.
For other factorial functions, see here.

I'll assume you want to find a closed form for $f(x)$ and then take the derivative. This is far easier and probably more of what you had in mind.
$$ f(x) = a \sum_{i=1}^x 10^{x-i}$$
$$ f(x) = a \sum_{i=1}^x 10^x/10^{i}$$
$$ f(x) = a \cdot 10^x \cdot \sum_{i=1}^x 1/10^{i}$$
$$ f(x) = a \cdot 10^x \cdot \left(\frac19 - \frac{1}{9 \cdot 10^x}\right) $$
$$ f(x) = a \left( \frac{10^x}{9} - \frac19 \right) $$
(I have used the closed form for a finite geometric series in evaluating the sum.)
Finding the derivative is now simple, it is:
$$ f'(x) = \frac{10^x a \log 10}{9} $$
