The electric potential and the electric field I'm preparing for my school exam in physics and I found that I don't understand the derivation of $E = -\nablaφ$. Here is a derivation from my book:

1) Imagine that 1 and 2 are infinitely close points with coordinates $x_1$ and $x_2$. They are so close that the electric field is a constant.
2) $dx = x_2 - x_1$
3) $-dφ = φ_1 - φ_2$
4) $Edx = -dφ \Rightarrow E = -\frac{dφ}{dx}$. Then we can generalize it on other coordinates (y and z) and be happy.

But I can't understand why $dφ$ does not equal $0$. It decreases more slowly than the electric field. So if the electric field is constant, the electric potential must be constant too.
 A: As far as I can tell, this "argument" is not a meaningful derivation of $\mathbf E = -\nabla \Phi$.  In step 4, one seems to assume that $Edx = -d\phi$ which is essentially what one is trying to prove in the first place, so the argument seems circular.
The precise derivation of this fact relies on the crucial fact that the electric field is a conservative vector field which then immediately guarantees the existence of a scalar function $\Phi$ for which $\mathbf E = -\nabla\Phi$.
As for your assertion

if electric field is a constant, electric potential must be a constant too.

Notice that if the electric field is constant in some region then the gradient of $\Phi$ is a constant there.  In particular its partial derivatives satisfy
$$
  \partial_x\Phi = -E_x, \qquad \partial_y\Phi = -E_y, \qquad \partial_z\Phi = -E_z
$$
The $x$ equation gives
$$
  \Phi(x,y,z) = -E_x x + f(y,z)
$$
Then the $y$ equation gives $\partial_yf = -E_y$ which implies $f(y,z) = -E_y y + g(z)$ and finally the $z$ equation gives $\partial_z g = -E_z$ so that $g(z) = -E_z z + c$.  Putting this all together gives
$$
  \Phi(x,y,z) = -\mathbf E \cdot \mathbf x + c
$$
In particular, the electric field being constant does not imply that the electric potential is constant, it means that it has the general form written in this last equation.
A: Since you are a high school student I'm going to make this as simple as possible. This should be read in addition to joshphysics's (correct) answer. The purpose of my own answer is to specifically fill in the steps that the book left out, but which are nevertheless correct.
First, I will take a guess that you have seen basic integral calculus and that your book has already defined the electric potential (in one dimension, for example $x$) in the following way:
$$\int_{x_1}^{x_2}E(x)dx=-\Phi(x_2)+\Phi(x_1).$$
For any well-behaved function
$$\lim_{\Delta a \rightarrow 0}\int_{a}^{a+\Delta a}f(x)dx=\lim_{\Delta a \rightarrow 0}f(a)\Delta a,$$
since we are evaluating the area of one thin rectangle with height $f(a)$ and width $\Delta a$. Anyway, this leads to, when $x_2$ is very close to $x_1$ (so that $x_2 = x_1 + \Delta x$ with $\Delta x$ becoming small):
$$\lim_{\Delta x \rightarrow 0}\int_{x_1}^{x_1+\Delta x}E(x)dx=\lim_{\Delta x \rightarrow 0}E(x_1)\Delta x=\lim_{\Delta x \rightarrow 0}\left(-\Phi(x_1+\Delta x)+\Phi(x_1)\right).$$
Continuing with this last step, we find
$$E(x_1)=\lim_{\Delta x \rightarrow 0}\left(-\frac{\Phi(x_1+\Delta x)-\Phi(x_1)}{\Delta x}\right)$$
or, since $x_1$ was arbitrary,
$$E(x)=\lim_{\Delta x \rightarrow 0}\left(-\frac{\Phi(x+\Delta x)-\Phi(x)}{\Delta x}\right)=-\frac{d\Phi(x)}{dx}.$$
The book is right that this can be extended to the variables $y$ and $z$.
