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In the geometric algebra formulation of Maxwell’s equation (singular in GA), the Green’s function for the spacetime derivative ends up with terms like

\begin{equation}\label{eqn:derivativeOfDeltaFunction:20}
\frac{d}{dr} \delta( -r/c + t – t’ ),
\end{equation}

or
\begin{equation}\label{eqn:derivativeOfDeltaFunction:40}
\frac{d}{dt} \delta( -r/c + t – t’ ),
\end{equation}

where \( t’ \) is the integration variable of the test function that the delta function will be applied to. If these were derivatives with respect to the integration variable, then we could use the well known formula

\begin{equation}\label{eqn:derivativeOfDeltaFunction:60}
\int_{-\infty}^\infty
\lr{ \frac{d}{dt’} \delta(t’) } \phi(t’) = -\phi'(0),
\end{equation}

which follows by chain rule, and an assumption that \( \phi(t’) \) is well behaved at the points at infinity. It’s not obvious to me that this can be applied to either of our delta function derivatives.

Let’s go back to square one, and figure out the meaning of these delta functions by their action on a test function. We wish to compute

\begin{equation}\label{eqn:derivativeOfDeltaFunction:80}
\int_{-\infty}^\infty \frac{d}{du} \delta( a u + b – t’ ) f(t’) dt’.
\end{equation}

Let’s start with a change of variables \( t” = a u + b – t’ \), for which we find

\begin{equation}\label{eqn:derivativeOfDeltaFunction:100}
\begin{aligned}
t’ &= a u + b – t” \\
dt” &= – dt’ \\
\frac{d}{du} &= \frac{dt”}{du} \frac{d}{dt”} = a \frac{d}{dt”}.
\end{aligned}
\end{equation}

Back substitution gives

\begin{equation}\label{eqn:derivativeOfDeltaFunction:120}
\begin{aligned}
a \int_{\infty}^{-\infty} \lr{ \frac{d}{dt”} \delta( t” ) } f( a u + b – t” ) (-dt”)
&=
a \int_{-\infty}^{\infty} \lr{ \frac{d}{dt”} \delta( t” ) } f( a u + b – t” ) dt” \\
&=
\evalrange{a \delta(t”) f( a u + b – t”)}{-\infty}{\infty}
–
a \int_{-\infty}^{\infty} \delta( t” ) \frac{d}{dt”} f( a u + b – t” ) dt” \\
&=
– \evalbar{ a \frac{d}{dt”} f( a u + b – t” ) }{t” = 0} \\
&=
\evalbar{ a \frac{d}{ds} f( s ) }{s = a u + b}.
\end{aligned}
\end{equation}

This shows that the action of the derivative of the delta function (with respect to a non-integration variable parameter \( u \)) is

\begin{equation}\label{eqn:derivativeOfDeltaFunction:160}
\boxed{
\frac{d}{du} \delta( a u + b – t’ )
=
\evalbar{a \frac{d}{ds}}{s = a u + b}.
}
\end{equation}