<s>[INST] Green's function and derivatives of Dirac's Delta In my textbook in a section about electromagnetism I came across the following lines: 
\begin{align}
\left[\left(\frac{\partial}{\partial x^0}\right)^2 - \Delta\right] \frac{\delta(x^0-|\mathbf{x}|)}{|\mathbf{x}|} &= \frac{1}{|\mathbf{x}|}\delta'' - \Delta \left(\frac{1}{|\mathbf{x}|}\right) \delta(x^0 - |\mathbf{x}|) - \frac{1}{|\mathbf{x}|} \Delta \delta(x^0-|\mathbf{x}|) - 2 \nabla \left(\frac{1}{|\mathbf{x}|}\right) \cdot \nabla \delta(x^0-|\mathbf{x}|)\\
&= \frac{1}{4\pi} \delta(\mathbf{x}) \delta(x^0-|\mathbf{x}|).
\end{align}
I'm having problems seeing the second equation; I know that
\begin{align}
\nabla^2 \frac{1}{|\mathbf{x}|} = -4\pi \delta(\mathbf{x}),
\end{align}
but I don't know how to deal with the derivatives of the delta distribution.
Can anyone explain, how to see that the equation above is correct?
Thanks in advance! 
 [/INST] Let's look at the a more general case:  a function of the form $f(x^0 - |\mathbf{x}|)/|\mathbf{x}|$.  For the sake of compactness, I'll switch to spherical polar coordinates where $|\mathbf{x}| = r$.  We therefore have
$$
\begin{multline}
\left[\left(\frac{\partial}{\partial x^0}\right)^2 - \nabla^2 \right] \frac{f(x^0-|\mathbf{x}|)}{|\mathbf{x}|} \\= \frac{f''(x^0 - r)}{r} - f(x^0 - r) \nabla^2 \left( \frac{1}{r} \right) - \frac{1}{r} \nabla^2 f(x^0 - r) - 2 \left[ \nabla \left(\frac{1}{r}\right) \right] \cdot \left[\nabla (f(x^0 - r)) \right]
\end{multline}
$$In spherical coordinates, we have
$$
\frac{1}{r} \nabla^2 f(x^0 - r) = \frac{1}{r} \left[ \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial f(x^0 - r)}{\partial r} \right)  \right] = \frac{1}{r} f''(x^0 - r) - \frac{2}{r^2} f'(x^0 - r)
$$
and 
$$
2 \left[ \nabla \left(\frac{1}{r}\right) \right] \cdot \left[\nabla (f(x^0 - r)) \right] = 2 \left( - \frac{1}{r^2}  \hat{r} \right) \cdot \left( - f'(x^0 - r) \hat{r} \right) = \frac{2}{r^2} f'(x^0 - r).
$$
(If you wish to check these, note that
$$
\frac{\partial f(x^0 - r)}{\partial r} = - f'(x^0 - r) \qquad \frac{\partial^2 f(x^0 - r)}{\partial r^2} = f''(x^0 - r)
$$
due to the chain rule.) Putting all of these together, we find that
$$
\frac{f''(x^0 - r)}{r}  - \frac{1}{r} \nabla^2 f(x^0 - r) - 2 \left[ \nabla \left(\frac{1}{r}\right) \right] \cdot \left[\nabla (f(x^0 - r)) \right] = 0,
$$
and so for any function $f$, we have
$$
\left[\left(\frac{\partial}{\partial x^0}\right)^2 - \nabla^2 \right] \frac{f(x^0-|\mathbf{x}|)}{|\mathbf{x}|} = - f(x^0 - r) \nabla^2 \left( \frac{1}{r} \right) = 4 \pi f(x^0 - |\mathbf{x}|) \delta^3(\mathbf{x}).
$$
The above derivation works for any function of $x^0 - r$ alone, including $\delta(x^0 - r)$.
The reason this works for any function, by the way, is that the general solution for the wave equation in three dimensions under the assumption of spherical symmetry (and excluding the origin) is
$$
u(r, t) = \frac{1}{r} \left[ f(r - t) + g(r + t) \right]
$$
for any two functions $f$ and $g$.
</s>