Is it true that $f(x)>g(x) \iff \frac {\operatorname d}{\operatorname d x }f(x)>\frac {\operatorname d}{\operatorname d x }g(x)$ ?
In one of the answers to this question this method was used to prove $f(x)>g(x)$.
PS : The origin of the question has additional conditions such as $f(a)=g(a)$ and $ a<x, \quad \frac {\operatorname d}{\operatorname d x }f(x)>\frac {\operatorname d}{\operatorname d x }g(x)$ then $a<x, \quad f(x)>g(x)$ (thanks to Donkey_2009)
No, neither implication is true.
$f(x) = 0, g(x) = -e^{-x}$. Then $f(x) > g(x) $ for all $x$, but $f'(x) < g'(x) $ for all $x$.
It is certainly not the case that $f(x)>g(x)$ at a point $x$ if and only if $f'(x)=g'(x)$ at $x$, since the derivative describes only local behaviour. It is not true either that if $f(x)>g(x)$ everywhere, then $f'(x)>g'(x)$ everywhere; indeed, nbubis's example of $f(x)=1+e^{-x}$ and $g(x)=1-e^{-x}$ satisfies $f>g$ everywhere and $f'<g'$ everywhere.
The result used in the linked question was the following:
Proposition Suppose $f(a)=g(a)$ and $f'(x)>g'(x)$ for all $x\ge a$. Then $f(x)>g(x)$ for all $x>a$.
Proof: Let $x>a$. Without loss of generality, $g(x)=0$ for all $x$ and $f(a)=0$. By the mean value theorem, we can write $f(x)=f(a)+(x-a)f'(\eta)=(x-a)f'(\eta)$, where $\eta\in[a,x]$. But $f'(\eta)\ge0$, and $(x-a)>0$, so $f(x)>0=g(x)$. $\Box$
Here's a descriptive argument. Considering $f(x)$ and $g(x)$ as the positions of two points at time $x$ then $f'(x)$ and $g'(x)$ are their velocities at time $x$. Now why in the world can one conclude $f(x)>g(x)$ from $f'(x)>g'(x)$?
Your statement is only true for monotonically increasing function(Think of derivative as a tangent to a curve).
For a decreasing(monotonic) function, the inequality reverses.
$$x>y \implies log_3{x}>log_3{y}$$
But,
$$x>y \implies log_{0.5}{x}<log_{0.5}{y}$$
