Electric field and potential Suppose we consider a point dipole i.e. an ideal dipole. At the origin the potential:
$$ \phi(\mathbf r) = \frac{1}{4\pi\epsilon_0} \frac{\mathbf p\cdot\hat{\mathbf r}}{r^2}$$
is undefined due to the $1/r^2$ term in the potential, but electric field:
$$ \mathbf E = \frac{1}{4\pi\epsilon_0}\frac{3(\mathbf p\cdot\hat{\mathbf r})\hat{\mathbf r} - \mathbf p}{r^3} - \frac{1}{3\epsilon_0}\mathbf p\delta^3(\mathbf r) $$
remains well defined. How can we calculate electric field at the origin by equation $\mathbf E=-\nabla\phi$?
 A: Both potential and electric field are singular at the point dipole, there is not much difference in "well-definedness" - there are no real values there. Electric field above is a singular distribution, not a function. It has no real value at the dipole. Only integrals of such distribution have value.
There are some fringe cases where electric field has value but isn't calculable as gradient of potential either. For example, electric field of charged sphere at any point on the sphere has definite value (half of electric field just above the sphere), but since potential is not differentiable at that point, electric field at such points is not a gradient of electric potential.
A: Just because a variable is zero at a certain position does not mean that the rate of change of that variable at the position is zero.  
When a ball is thrown up, at its greatest height the ball’s vertical velocity is zero but its rate of change of velocity, the acceleration of free fall, is not zero.  
In your case the potential at a distance $x$ from a charge $+q$ is $\phi=\frac{k(+q)}{x}+\frac{k(-q)}{d-x}$ where $d$ is the separation of the charges.  
When $x=\frac d2$ the potential $\phi$ is zero but the electric field (minus the gradient of the potential) is non zero.
