Let's only work with functions $f(x)$ that have a series expansion at $x=0$. Is it true that: $$ {d O(1)\over d x} = O(1) $$ for all such functions $f(x)$? Here $O$ is the big-O notation and we are expanding around the point $x=0$ (i.e. not around $x\to\infty$).
The only counter examples that I can find are of the type $f(x)=x^n\sin(1/x)$ where $n$ can be 0, 1, 2, ...., but these functions do not have a series expansion around $x=0$, so I am not interested in them.
I think the question really boils down to what exactly you mean by "has a series expansion". As I noted in a comment, $O(1)$ at $0$ just means "bounded near $0$". Depending on how you define "has a series expansion", this may be true by definition. For instance, if you require that $f'(x)$ to exist and be continuous near $x=0$, then the result you want becomes trivially true (being continuous locally always implies being bounded locally, just by the $\epsilon$, $\delta$ definition of continuity).
There may be some issues with extended definitions of "having a series expansion," though (I'm not sure). For instance, I'm not sure what happens if you allow series expansions like $a_{1/2}x^{1/2} + a_{3/2}x^{3/2} + ...$.