# Griffiths Electrodynamics gradient of charge distribution

 P: 166 I do not understand the following from Griffiths’ Electrodynamics – page 424 Equation 10.21. $$\nabla p = \dot{p} \nabla {tr} = …$$ I’m not sure how much of this applies (I think my question is on the math) but p is the charge distribution, tr is the retarded time. Is this an application of the chain rule? With the gradient being a derivative with respect to spatial location (x,y,z), why is the time derivative showing up in the gradient? I initially want to say if something is dependent upon t but not on x, then its derivative with respect to x is zero. The result looks like the chain rule applied – I don’t see why the time dependent portion shows up. Can you help clear this up for me? Thanks Sparky_
 P: 166 I do not see it yet. I see later on the same page $$\nabla \dot{p} = \ddot{p} \nabla {tr} = …$$ Can you explain further? Somehow the gradient is giving an additional time derivative. Thanks Sparky_
 P: 584 Griffiths Electrodynamics gradient of charge distribution $\rho$ has arguments like this: $\rho (\vec{r}', t_r(\vec{r}, \vec{r}', t))$ The gradient is being applied w.r.t to the coordinates of $\vec{r}$ ( not $\vec{r}'$ which gets integrated away). The coordinates that we would be taking the derivative with respect to in order to obtain the gradient are only found in the parameters of $t_r$. So this result is from the chain rule. Here is one component of the gradient, for example. $(\nabla \rho)_x = \frac{\partial \rho (\vec{r}', t_r(x, y, z, \vec{r}', t))}{\partial x} = \dot{\rho}\frac{\partial t_r}{\partial x}$