Big-O notation Basics, is it related to derivatives?

Big-Oh is is not completely determined by derivatives. For example $\sin(x^2)\in O(1)$ but the derivative $2x\cos(x^2)$ is unbounded.

The claim that $f(n)\le g(n)$ implies $f(n)\in O(g(n))$ is false: Consider $g(n)=n$, $f(n)=-n^2$. But if you replace the condition with $|f(n)|\le g(n)$ then the claim is easy: That is almost the definition of $f(n)\in O(g(n))$. And of course trivially $g(n)\in O(g(n))$. Then since Big-ohs are closed under addition, also $f(n)+g(n)\in O(g(n)$.

I found this question (and the first answer) helpful: Big-O Notation and Asymptotics

For example, $f(n)$ is $O(g(n))$. Then, $f(n)$ may diverge (increase without bound). However, $(f(n))/(g(n))$ does not, as $g(n)$ is always greater than $f(n)$ beyond some number $N.$

So, really, it has more to do with the limit of the ratio of two functions than derivatives.

The definition of the derivative can be expressed using asymptotic notation.

We say f has a derivative at x if there exists M such that:

$$f(x+\epsilon) = f(x) + M\epsilon + o(\epsilon)$$

We denote this M as f '(x)

(edited as per Antonio's correction)