How would you take the derivative of this function? $$f(x)=\sum_{i=1}^x\frac1i$$
$$f'(x)=?$$
NOTE: I am defining f(x) for all natural number and considering the function as you zoom out
Is it even possible to take this derivative? Why or why not? If so, what meaning does the derivative convey? What function does f(x) approach as you zoom out? These are all questions that popped into my head when considering this function and now I look to you to answer them because my small brain can not.
 A: $f(x)$ has a natural extension to complex $x$ as the harmonic number function
$$H(x)=\psi(x+1)+\gamma$$
Here $\psi$ is the digamma function. It follows that
$$H'(x)=\psi^{(1)}(x+1)$$
with a trigamma function.
A: It might interest you to know that the premise for what you are attempting to do, for a general $f(i)$ not just
for the particular case of $f(i)=\frac{1}{i}$, goes under the name of
Indefinite_Sum
or "Anti-Delta"..
In your particular case, since the Digamma function
satisfies the functional identity
$$
\Delta _x \psi (x) = \psi (x + 1) - \psi (x) = \frac{1}{x}
$$
then
$$
\sum\limits_x {\frac{1}{x}}  = \Delta _x ^{ - 1} \left( {\frac{1}{x}} \right) = \psi (x)
$$
Therefore
$$
f(x) = \sum\limits_{k = 1}^x {\frac{1}{k}}  = \sum\nolimits_{k = 1}^{x + 1} {\frac{1}{k}}  = \psi (x + 1) - \psi (1)
$$
note that the second sum is the correct translation of the first under the definition of the indefinite sum,
when this is made "definite".
Consequently
$$
f'(x) = \psi ^{\left( 1 \right)} (x + 1)
$$
