Delta dirac gradient in Liénard-Wiechert fields derivation I'm following a derivation of the Lienard-Wiechert field in Zanwill, Modern Electrodynamics, which involves de gradient of a Dirac delta function.
EDIT:
The Lienard-Wiechert problem is to find the potentials and fields of a moving point charge.
The result for the potentials is
\begin{equation}
\phi(\mathbf r, t) = \dfrac{q}{4\pi \epsilon_0} \int dt'\dfrac{\delta(t'-t+R(t')/c)}{R(t')} 
\end{equation}
\begin{equation}
\mathbf A(\mathbf r, t) = \dfrac{\mu_0 q}{4\pi} \int dt'\dfrac{\mathbf v(t')\delta(t'-t+R(t')/c)}{R(t')}
\end{equation}
Where $R(t)=\|\mathbf R(t)\|=\|\mathbf r-\mathbf r'(t)\|$, being $\mathbf r$ the observation point, $\mathbf r'(t)$ the position of the point charge, and, $t'$ is a dummy integration variable.

Now, in particular, for the electric field $\mathbf E=-\nabla\phi-\partial_t \mathbf A$ ($\leftarrow$ gradient with respect to $\mathbf r$)
\begin{equation}\mathbf E(\mathbf r, t)=-\dfrac{q}{4\pi \epsilon_0} \nabla\int dt'\dfrac{\delta(t'-t+R(t')/c)}{R(t')} - \dfrac{\mu_0 q}{4\pi} \partial_t\int dt'\dfrac{\mathbf v(t')\delta(t'-t+R(t')/c)}{R(t')}\end{equation}
In the next line, it says that just using the property $\nabla R=\mathbf{\hat n}$ and the chain rules gives
$$\nabla\delta(t'-t+R(t')/c)=-\partial_t\delta(t'-t+R(t')/c)\,\frac{\mathbf{\hat n}}{c}$$
That's exactly the step I don't understand quite well.
It is easy to identify the term $\dfrac{\mathbf{\hat n}}{c}$ as the gradient of the function inside the $\delta$ function. And it gives a sense of having used this property,
\begin{align}
\nabla_{\mathbf x}\delta(\mathbf x-\mathbf x')=-\nabla_{\mathbf x'}\delta(\mathbf x-\mathbf x')
\end{align}
with $\nabla_{\mathbf x} \rightarrow \nabla$ and $\nabla_{\mathbf x'} \rightarrow \partial_t$. But, as I said, it is not quite clear.
 A: Since the Dirac delta is a tempered distribution, it has well-defined derivatives. The Dirac delta in question is a function of a time variable, but the time argument has some radial spatial dependence.
For convenience of notation, let
\begin{equation}
\tau(t,t') = t' - t + \frac{R(t')}{c}.
\end{equation}
Then we are interested in
\begin{equation}
\begin{split}
\nabla \delta\left(\tau\right)
&=~
\hat{\mathbf{e}}_R\frac{\partial}{\partial R}\delta\left(\tau\right)~~~~~(\textrm{No angular dependence})\\
&=~
\hat{\mathbf{e}}_R\frac{\partial\tau}{\partial R}\frac{\partial}{\partial\tau}\delta\left(\tau\right).
\end{split}
\end{equation}
Now note that
\begin{equation}
\begin{split}
\frac{\partial\tau}{\partial R} &=~ \frac{\partial}{\partial r}\left(t' - t + \frac{R(t')}{c}\right)\\
&=~ \frac{\partial}{\partial R}\left(\frac{R(t')}{c}\right)\\
&=~ \frac{1}{c}.
\end{split}
\end{equation}
We can now "re-cast" $\partial/\partial\tau$ in terms of $\partial/\partial t'$ or $\partial/\partial t$. Since the pictured integration is in $t'$, further integration will be in $t$.
\begin{equation}
\begin{split}
\frac{\partial}{\partial\tau}\delta(\tau) &=~
\frac{\partial t}{\partial\tau}\frac{\partial}{\partial t}\delta\left(t' - t + \frac{R(t')}{c}\right)\\
&=~ -\frac{\partial}{\partial t}\delta\left(t' - t + \frac{R(t')}{c}\right)
\end{split}
\end{equation}
because
\begin{equation}
t = t' - \tau + \frac{R(t')}{c}
\end{equation}
or
\begin{equation}
\frac{\partial t}{\partial\tau}
 = \left(\frac{\partial\tau}{\partial t}\right)^{-1} = -1.
\end{equation}
