Field of a Polarized Object In Griffith's Electrodynamics, in the section 4.2, just after the equation 4.9, he writes "sleight-of-hand casts this integral into a much more illuminating form"...
I have a doubt in that. If the Gradient (or differentiation if carried out) is with respect to primed coordinates, how can variable r be differentiated as r' ? It would be of great help if someone clarifies this point.
 A: The point is that the 'scripty r' (i dont know how to write it here)  depends only on the difference between the coordinates; note that
($\frac{\partial}{\partial x}$) $f(x- x')$ = -($\frac{\partial}{\partial x'}$)$f(x- x')$. What i am trying to say is that the 'scripty r' is a vector joining the primed co-ordinates to the unprimed co-ordinates.....hence the gradient is taken wrt either of the co-ordinates....ie, gradient wrt unprimed is negative of gradient wrt primed co-ordinates
A: We need to shift from $\mathscr R$ to $r'$ because otherwise the coordinate system would keep changing as we integrate over the whole volume. Now, about that sleight of hand.
Gradient depends upon the coordinate system.
By simple definition of gradient we have :-
$dT= \nabla T.\boldsymbol{dl}$, where $\boldsymbol{dl}$ is the change in space of the coordinate system.
Since, $\mathscr R = \boldsymbol r - \boldsymbol r' $ it implies that $d\mathscr R = -d\boldsymbol r'$ as $\boldsymbol r$ is the constant position vector of the point of interest where we wish to calculate electric field by the polarized object, in the source coordinate system.
Now, $d \left( {\frac {1}{\mathscr R}} \right) = \nabla \left( {\frac {1}{\mathscr R}} \right).d\mathscr R = \nabla'\left( {\frac {1}{\mathscr R}} \right).d\boldsymbol r'$, as $d\mathscr R = -d\boldsymbol r'$ this implies
$\nabla' \left( {\frac {1}{\mathscr R}} \right)$ = $-\nabla \left( {\frac {1}{\mathscr R}} \right)$ which simply means that gradient in source coordinate system is just the negative  of the gradient in the coordinate system of that differential dipole in consideration, which is what Griffiths touched upon.
PS : When I was having trouble with this, I was incorrectly assuming $d|\mathscr R|  \hat{ \mathscr R} = -dr'  \hat{r'} $, this was incorrect because I was incorrectly assuming $d\hat{ \mathscr R} = d\hat{r'}= 0$
