How to take time derivative of gradient in deriving the retarded scalar potential? In Griffiths E&M 4th edition, equation 10.28 reads
$$\nabla \rho = \dot{\rho}\nabla t_r$$
This is in the context of computing the gradient of the retarded potential
$$V(\vec{r'},tr)=\frac{1}{4\pi \epsilon_0}\int \frac{\rho(\vec{r'},t_r)}{\mathit{r}_1}d\tau'$$ where
$$t_r \equiv t-\frac{r}{c}\\
\mathit{r}_1 \equiv |\vec{r}-\vec{r'}|.$$
The step is clear, using the product rule for the gradient
$$\nabla V= \frac{1}{4\pi \epsilon_0}\int\left[\left(\nabla \rho \right )\frac{1}{\mathit{r}_1} + \rho \nabla \left(\frac{1}{\mathit{r}_1} \right ) \right ]d\tau'.$$
Then the problem I am having is understanding the next step. Without explanation or reference, states that
$$\nabla \rho = \dot{\rho}\nabla t_r.$$
I can follow the rest of it, but I don't see how he goes from $\nabla \rho(r,t_r)$ to $\dot{\rho}\nabla t_r$.
It seems like a small matter in such a complex subject, but I could use some help explaining this.
BTW I get that $\dot{\rho}$ is the time derivative of $\rho$.
 A: I had a brief look at Griffiths. Not sure what $\int\dots d\tau'$ is, but it looks like integration with respect to $\mathbf{r}'$, i.e. the position of the source ($\int\dots d\tau'=\int\dots d^3r'$). Which would make sense.
So then you have:
$$
\boldsymbol{\nabla}\int d^3 r' \dots\rho\left(\mathbf{r}',\,t_r \right)=\int d^3 r' \dots \boldsymbol{\nabla}\rho\left(\mathbf{r}',\,t_r \right)
$$
Now you have to understand $\boldsymbol{\nabla}$ is - it is derivative with respect to $\mathbf{r}$ - the position of the observer. That position is independent from what of the source, i.e. $\boldsymbol{\nabla}r'_i=0$ for all $i$. The only way $\rho\left(\mathbf{r}',\,t_r \right)$ depends on the position of the observer is through the retarded time. Hence, by chain rule:
$$
\boldsymbol{\nabla}\rho\left(\mathbf{r}',\,t_r \right)=\frac{\partial \rho\left(\mathbf{r}',\,t\right)}{\partial t}\Bigg|_{t=t_r}\cdot\boldsymbol{\nabla}t_r
$$
Finally, Griffiths uses notation:
$$
\frac{\partial \rho\left(\mathbf{r}',\,t\right)}{\partial t}\Bigg|_{t=t_r}=\dot{\rho}\left(\mathbf{r}',\,t_r\right)
$$
