Does $f(x)>g(x)$ imply $\frac{d}{dx}f(x)>\frac{d}{dx}g(x)$? Is it true that $f(x)>g(x) \implies \frac{d}{dx}f(x)>\frac{d}{dx}g(x)$?
What about $|f(x)|>|g(x)| \implies \frac{d}{dx}|f(x)|>\frac{d}{dx}|g(x)|$?
 A: Neither of these statements are true in general. Simple counter-example for both: $f=1$, $g=0$.
A: Hint: Consider $f(x)=\pi$ and $g(x)=\arctan(x)+\pi/2$.
A: This is generally not true. Consider $f(x)=ax+b$ and $g(x)=ax+b+\frac {1} {x}$ with $a,b > 0$. Then
$$ f(x)=|f(x)| < g(x)=|g(x)| $$
with 
 $$f'(x)=a > g'(x)=a-\frac {1} {x^2}$$
So 
$$f(x) < g(x) \text{ does not imply } f'(x)<g'(x) $$
However  $$\text{ If } f'(x) > g'(x) \text { and } f(c)\geq g(c)  \text{ for some $c \in \mathbb{R}$ then  } f(x)>g(x) \text { for $x > c$ }$$
is true for differentiable functions
A: No and I belive that the simplest way to see this is considering a positive decreasing function on $\mathbb{R}$ like taking $f(x)=e^{-x}$ and $g(x)=0$. We have $$\forall x\in\mathbb{R},\,f(x)>g(x)$$ but for any $x\in\mathbb{R}$ we have $f'(x)=-e^{-x}<0$ and $g'(x)=0$ and so $$\forall x\in\mathbb{R},\,f'(x)<g'(x)$$
Also to not believe that then $f(x)>g(x)\Rightarrow f'(x)<g'(x)$ you can take a positive function that is positive and increasing in some interval and decreasing in another one. Take for example $g=0$ and $f(x)=x^2+1$. Again: $f>g$ but $$\forall x\in\mathbb{R}^-,\,f'(x)<g'(x)$$ and $$\forall x\in\mathbb{R}^+,\,f'(x)>g'(x)$$ and $$f'(0)=g'(0)$$ so that neither $f>g$ nor $f<g$ are true.
