Why does ∇'(1/X) equal x/X² in Griffiths derivation?

[nuclear_dog]

In Griffiths, for deriving the bound charges for a given polarization P , the formula used is the general formula for dipoles .i.e ( equation 4.9)
{Here the potential at r is calculated due to the dipole at r' )

V(r) = ∫$\frac{x.P(r')}{X^2}$d$\tau'$

Here X = r - r' , and x = unit vector in the direction of X

Then it is written that $\frac{x}{X^2}$ = $\nabla'$(1/X).

since X = (r-r') , and ∇' = (∂/∂r')$\widehat{r'}$ ...

Shouldn't ∇'(1/X) be (1/X^2)$\widehat{r'}$ ?

 

[TSny]

It might be best to work it out in Cartesian coordinates where all coordinates are written explicitly. See if you can fill in the details of the derivation outlined in the figure.

 

[nuclear_dog]

Thanks , I can see that in Cartesian coordinates .