Whats the difference between differentiating w.r.t Source Coordinates and differentiating w.r.t Field Coordinates? While reading the bound charges section on Griffith(ED), I came upon the equation:
$\vec{\nabla}^\prime\bigg({1\over r}\bigg)={\hat{r}\over r^2}$
And Griffith goes onto say that the prime differentiation is different from the usual one. ie.,
$\vec{\nabla}\bigg({1\over r}\bigg)={\hat{r}\over r^2}$
So can someone please explain what is the difference between the two?  
Thanks

 A: Actually, the two gradients should differ by sign.
You may think of $\frac{1}{r}$ as a function of variables $x, y, z, x^{\prime}, y^{\prime}, z^{\prime}$.
In this case, we can write the function as
$$
f = \frac{1}{\sqrt{(x-x')^{2} + (y-y')^{2} + (z-z')^{2}}}
$$
Then, we have that $$
\frac{\partial f}{\partial x} = -\frac{x-x'}{[(x-x')^{2} + (y-y')^{2} + (z-z')^{2}]^{\frac{3}{2}}}
$$ 
and that 
$$
\frac{\partial f}{\partial x'} = \frac{x-x'}{[(x-x')^{2} + (y-y')^{2} + (z-z')^{2}]^{\frac{3}{2}}}
$$
Use similar relations for the other variables and you get the two gradients, which should be identical except for different signs, and then just rewrite in terms of r to get what Griffiths has in the book.
A: $\textbf{r}$ are the points where you evaluate the field. 
$\textbf{r'}$ are the points where you have your charges or currents.
A prime gradient indicates you have to derivate by the prime variables $x'$,$y'$,$z'$:
$\nabla' f(x,y,z)=\left(\frac{\partial f(x,y,z)}{dx'},\frac{\partial f(x,y,z)}{dy'},\frac{\partial f(x,y,z)}{dz'}\right)$
