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.