Is Feynman correct about this inverse square relationship in Vol I-28? In the Feynman Lectures, Volume I, 28-1, Feynman promulgates this
expression for the electric field due to a moving charge. My notation
differs from the original.
$$
\mathfrak{E}=-\frac{q}{4\pi\varepsilon_{o}}\left(\frac{\hat{\rho}}{\rho^{2}}+\frac{\rho}{c}\frac{d}{dt}\left[\frac{\hat{\rho}}{\rho^{2}}\right]+\frac{1}{c^{2}}\frac{d^{2}\hat{\rho}}{dt^{2}}\right).
$$
Here $\rho$ represents the retarded position of the source; and $\hat{\rho}$
is the corresponding unit vector point away from the source.
He later asserts that 

Of the terms appearing in [this equation],
  the first one evidently goes inversely as the square of the distance,
  and the second is only a correction for delay, so it is easy to show
  that both of them vary inversely as the square of the distance.

The first term is simply Coulomb's inverse square law:
$$
\mathfrak{E}_{1}=-\frac{q}{\rho^{2}4\pi\varepsilon_{o}}\hat{\rho}.
$$
Representing the delay $\frac{\rho}{c}=\Delta t$, the second term
can be written as
$$
\mathfrak{E}_{2}=\frac{d}{dt}\left[-\frac{q}{\rho^{2}4\pi\varepsilon_{o}}\hat{\rho}\right]\Delta t.
$$
And it is indeed the time derivative of the first term (multiplied
by the delay): 
$$
\mathfrak{E}_{2}=\frac{d\mathfrak{E}_{1}}{dt}\Delta t.
$$
The third term is not dependent on $\rho$:
$$
\mathfrak{E}_{3}=-\frac{q}{4\pi\varepsilon_{o}c^{2}}\frac{d^{2}\hat{\rho}}{dt^{2}}.
$$
Carrying out the differentiation in the second term gives
$$
\mathfrak{E}_{2}=-\frac{q}{4\pi\varepsilon_{o}c}\rho\left(\frac{1}{\rho^{2}}\frac{d\hat{\rho}}{dt}-\frac{d\rho}{dt}\frac{2}{\rho^{3}}\hat{\rho}\right)
$$
$$
=-\frac{q}{4\pi\varepsilon_{o}c}\left(\frac{1}{\rho}\frac{d\hat{\rho}}{dt}-\frac{d\rho}{dt}\frac{2}{\rho^{2}}\hat{\rho}\right).
$$
The first term in the parentheses varies as the inverse of
$\rho$. The second term does vary as the inverse squared. 
I contend Feynman is incorrect. Is he?
I'm adding this graphic to illustrate the answer I received.  I was, in fact, wrong.

 A: I would suggest that you reconsider the quantity $d\hat{\rho}/dt$, which itself varies inversely with $\rho$.
Now, $\hat{\rho}$ is the unit vector pointing from the field point $P$ at which $\mathfrak{E}$ is measured to the charge $q$.  At this point, let's introduce a new variable that Feynman evidently didn't want to introduce: the velocity $\vec{v}$ of the charge.  You'll surely agree that this is independent of the field point $P$.
It also helps to think of a few specific scenarios, and consider how the result would change.  So imagine that $\vec{v}$ is perpendicular to $\hat\rho$ and has magnitude $10\mathrm{m/s}$.  If $\rho$ is just a few meters, you expect $d\hat{\rho}/dt$ to be quite large.  On the other hand, if $\rho$ is several light years, you expect $d\hat{\rho}/dt$ to be extremely small.  So you might begin to believe that it varies inversely with $\rho$.
But to be a little more specific, the only possible change in $\hat{\rho}$ is a rotation, and can therefore be described by an angular velocity $\vec{\omega}$.  This angular velocity is just the angular velocity  of the charge as seen from the field point.  A standard formula gives the angular velocity as
\begin{equation}
  \vec{\omega} = \frac{\hat{\rho} \times \vec{v}} {\rho},
\end{equation}
so we have
\begin{equation}
  \frac{d \hat{\rho}} {dt} = \vec{\omega} \times \hat{\rho} = \frac{(\hat{\rho} \times \vec{v}) \times \hat{\rho}} {\rho}.
\end{equation}
(Note that the cross-product is not associative, so the grouping by parentheses is crucial.  In particular, the numerator of that final expression is not trivially zero.)  This does indeed vary inversely with $\rho$, which means that your term $\frac{1}{\rho} \frac{d \hat{\rho}} {dt}$ varies inversely with the square of $\rho$, just as Feynman claimed.
