Is the derivative of a function bigger or equal to $e^x$ will always be bigger or equal to the function ?! It seems to be the case, but i don't have a proof.
Given the function $f$ such that $f(x) \geq e^x$, is it true that $f'(x) \geq f(x)$?!
I was experimenting with wolfram and it appears that $\frac{f'(x)}{f(x)} \geq 1$ whenever $f$ is bigger or equal to $\exp(x)$. 
Note : as suggested in the comments, i meant that for all positive $x$ which means that $f(x) \geq e^x \space \forall x$ such that $x\ge 0$.
 A: No and heres why.
Assume u have a function $f(x) = e^x+1$ which is greater than $e^x$ for all $x$.
However its derivative would be $f'(x) = e^x$ which is smaller. I think a more interesting question would be if the function was asymptotically greater than $e^x$
A: Let us consider the function $g(x):=\log(f(x))$.
Your question is recast as: if $g(x)\ge x$, is it true that $g'(x)\ge1$ ?* 
This is obviously false, as a curve lying above $x$ can take any slope.


*$\log f(x)\ge \log e^x$ vs. $(\log f(x))'=\dfrac{f'(x)}{f(x)}\ge1$.
A: This is not true. The $e^x$ is always positive, so it suffices to give a function that is decreasing on some interval and yet greater than $e^x$. It should be obvious that such functions exist.
A: let $f(x) = e^x + 1 + \cos x \ge e^x + 1 - 1 = e^x$
$f'(x) = e^x - \sin x$. 
A: If $f(x) \geq e^x \space \forall x$ then we can take the logarithm of both sides to give $\log f(x) \geq x \space \forall x$. It is tempting to take the derivative of both sides and conclude that $\frac{f'(x)}{f(x)} \geq 1 \space \forall x$ - but, as the examples above show, that is just not true.
