Gradient of function in terms of retarded time produces time derivative

Question

In Griffith's introduction to electromagnetism on page 479, there is this following equation:

\begin{align*} \nabla V &\cong \nabla \left[\frac{1}{4\pi \epsilon_0} \frac{\boldsymbol{\hat r} \cdot \dot{\mathbf{p}}(t_0)}{rc} \right] \\ &\cong \left[\frac{1}{4\pi \epsilon_0} \frac{\boldsymbol{\hat r} \cdot \ddot{\mathbf{p}}(t_0)}{rc} \right] \nabla t_0 \\ &= -\frac{1}{4\pi \epsilon_0 c^2} \frac{[\boldsymbol{\hat r} \cdot \ddot{\mathbf{p}}(t_0)]}{r} \boldsymbol{\hat{r}} \end{align*}

given that $t_0\equiv t - \dfrac{r}{c}$. Why is it that one the second line the dipole moment -- the $p$ function -- gains an extra time derivative if the gradient is in terms of partial spatial derivatives?

Answer 1

$t_0=t-r/c$ is a function of $r$, so the $\nabla$ acting on it is the same as $\partial_t$ (complicated a bit by signs and $c$). It would be easier to see if he didn't use $t_0$, and put it all in terms of $(t-r/c)$ explicitly.