Prove that $f'(x_0)=c$ 
Let $f:(a,b)\rightarrow \mathbb{R}$, and $x_0 \in (a,b)$. $f$ is differentiable at $(a,b)$. Also, Let $l(x)= cx+d$, "passes" at $(x_0, f(x_0))$.
  Prove that if $\forall x \in (a,b):f(x) \ge l(x)$ then $f'(x_0)=c$.

It looked suitable using the derivative definition here:  
$$f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f({x_0} + h) - f({x_0})}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{f({x_0} + h) - c{x_0} - d}}{h}$$
But how to continue from here? (If that's the desirable way)  
EDIT:

Maybe defining $g(x)=f(x)-l(x)$ is the right way to the conclusion?  
The derivative is: $g'(x) = f'(x)-l'(x) = f'(x)-c$.
Comparing the derivative to $0$ we get:
$f'(x)-c = 0$. Therefore, $f'(x)=c$.  
So we know that when $g'(x)=0$, then $f'(x)=c$, but what can we learn from it?

Your idea of defining $g(x)$ is correct. Since $g(x)\ge 0$ by the property $f(x)\ge l(x)$ you have that the minimum of $g$ is at $x_0$ since $$g(x_0)=f(x_0)-l(x_0)=0$$ So, since there is the minimum you have that $g'(x_0)=0$ (by the first derivative criterion for a minimum of a differentiable function). But this means that $$0=g'(x_0)=f'(x_0)-l'(x_0)=f'(x_0)-c$$ which implies that $f'(x_0)=c$.
