prove , if $ \frac {dy}{dx} = 0 $ then $ y = c$ for some constant $c$ . we all know that if $ y = c $ for some constant $c$ then , $ \frac{dy}{dx} = 0 $
but how can we prove the other way ? , i mean , how can we 
prove that , if $ \frac {dy}{dx} = 0 $ then $ y = c$ for some constant $c$ .
that is not a homework . it's just a question i want to know its answer ! 
i hope that you take my question seriously :) 
so , can anyone give us the simplest prove ?!! 
thanx for all 
 A: Just use the Mean-Value Theorem. For any $ x,y \in \mathbb{R} $ satisfying $ x < y $, we have
$$
f(x) - f(y) = f'(c) \cdot (x - y)
$$
for some $ c \in (x,y) $. As $ f' \equiv 0 $, it follows that $ f(x) = f(y) $.
A: Assume $f'(x) = 0$ for all $x\in \mathbb{R}.$
By the Mean-Value Theorem, for any $ x,y \in \mathbb{R} $ such that $ x < y $, we have
$$f(x) - f(y) = f'(c) (x - y)$$
where $c \in (x,y)$. Since $ f' = 0,\;$ by assumption, it follows that $ f(x) - f(y) = 0\cdot (x-y$ \implies f(x) = f(y)$.
But $ f(x) = f(y) \implies f(x) = f(y) = C$, for some constant C. That is, $f(x) = C$ is the constant function.

Intuitively, you can think of $dy/dx = f'(x)$ as the slope of a line, where $f'(x) = 0$ implies that the line is horizonal, i.e. the corresponding equation is given by $y = 0\cdot x + c \implies y = f(x) = c$, a line parallel to the x-axis and intersecting the y-axis at $c$.
A: The answer depends on how you rigorously define your notion of differentiation. If you go the route Andres has suggested, you are using the geometric definition of differentiation.
On the other hand, the algebraic definition of differentiation (see http://en.wikipedia.org/wiki/Formal_derivative for instance) takes your statement as an axiom allows your statement to be proved in a more general context.
A: If $f'=0$, then for every $x,y$ we have $\displaystyle f(y)=f(x)+ \int_x^yf'(t)dt=f(x)$. Therefore, $f$ is constant.
