Big-O notation Basics, is it related to derivatives? I am having the hardest time with Big-O notation (I am using this Rosen book  for the class I am in). 
On the surface, Big-O reminds me of derivatives, rate of change and what not; is this proper thinking? If $f(n)$ is $O(g(n))$, would the derivatives have any affect on this?
Essentially is there a good resource for learning Big-O for the first time?
If I missunderstand this forum and need a specific question, then:
Prove that if $f(n)\le g(n)$ for all $n$, then $f(n) + g(n)$ is $O(g(n))$. (I'd rather gain an understanding of how to do this than to have an answer to a problem).
EDIT:
My attempt at the answer to my specific question using l'Hôpital:
$$\lim_{x\to\infty} \frac{f'(x)}{f'(x) + g'(x)} = \lim_{x\to\infty} \frac{1}{g'(x)}.$$
 A: 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 $f(n)+g(n)\in O(g(n)$.
A: 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.
A: 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)
A: Concrete Mathematics is a good place to start with asymptotics and related ideas, particularly if you are in computer science (which your tags suggest).
A: One more easy way of looking at this type of problem is noticing that if $f(n) \leq g(n)$ then
$$
f(n)+g(n) \leq g(n) +g(n)=2g(n)= O(g(n))
$$
In fact it is easy to show that this sum is $\Theta(g(n))$
