Help me understand eqn. 28.6 from Feynman Feynman says that the electric field, $\bf{E}$, can be written as,
$$\mathbf{E} = \frac{-q}{4 \pi \epsilon_0} \left[ \frac{ \mathbf{e}_{r'}}{r'^2} + \frac{r'}{c} \frac{d}{dt} \left(\frac{\mathbf{e}_{r'} }{r'^2}\right) + \frac{1}{c^2} \frac{d^2}{dt^2} \mathbf{e}_{r'} \right]$$
where $\mathbf{E}$ is the electric field at a point, P, from a charge, $q$, that is a distance $r$ apart. My impression is that $\mathbf{e}_{r'}$ is the unit vector from P in the direction of $q$, presumably $\frac{r'}{||r'||}$.
The first of the three terms is Coulomb's law, the second term is an apparent time-delay Coulomb field, and the third is the predominant factor contributing to the electric field for a large distance $r$.
Since radiation behaves as if it were inversely related to the distance and the distance is large, the first two terms are ignored leaving us with,
$$\mathbf{E} = \frac{-q}{4 \pi \epsilon_0 c^2 } \frac{d^2 \mathbf{e}_{r'}}{dt^2}.$$
A further simplification is sought for when "charges are moving only a small distance at a relatively slow rate." Now he argues that the charge's motion is so tiny that its distance does not change and so the relay remains $r/c$. Now he says,  


Then our rule becomes the following: If the charged object is moving in a very small motion and it is laterally displaced by the distance $x(t)$, then the angle that the unit vector $\mathbf{e}_{r'}$ is displaced is $x/r$, and since $r$ is practically constant, the $x$-component of $d^2 \mathbf{e}_{r'}/dt^2$ is simply the acceleration of $x$ itself at an earlier time, and so finally we get the law we want, which is
    $$\mathbf{E}_x(t) = \frac{-q}{4 \pi \epsilon_0 c^2 r}a_x \left(t - \frac{r}{c}\right).$$


See here for the relevant portion. My problem is that I do not understand how he arrived at this formula. How I picture this scenario is what I've drawn: 
I picture a unit vector for $e_{r'}$ sitting at the dark spot, point P or where we measure E, along the line of $r'$ and a corresponding one for $r$. The angle between the two unit vectors is small, so that $\sin \theta = \frac{x}{r} $, but the angle is small enough that the approximation can be used to give $\theta = \frac{x}{r}$.
What I have trouble now is making a formula of the form $\mathbf{e}_{r'}(t)$. I do think that $\mathbf{e}_r' = \frac{r'}{||r'||}$ makes sense as a starting point, but what I need to describe is the time-dependent behavior. It seems like $r = \sqrt{r'^2 + x^2}$ is also equally valid to write. We might say that $r(t) = \sqrt{ r'^2 + x(t)^2}$, where $x(t) = \dot{x} t$. Now writing, $r(t) = \sqrt{ r'^2 + \dot{x}t}$. So say I want to compute $\frac{d e'_x}{dt} =\lim_{h \rightarrow 0}  \frac{\frac{r(t+h) - r(t)}{h}}{||r'||} = \lim_{h \rightarrow 0}  \frac{\frac{\sqrt{r'^2 + \dot{x}(t+h)}) - \sqrt{r'^2 + \dot{x}(t)}}{h}}{||r'||}$, but this appears to be getting pretty messy already and I haven't even tackled the second derivative.
What am I missing here?
EDIT: In RE Art Brown Answer
The following image reflects my understanding of your description:

Effectively we ignore any $y$ component of $\mathbf{e}_r$, but we still need to express the $x$-component of $\mathbf{e}_r$ before we can calculate its second derivative. Looking at the top view it seems that $\mathbf{e}_r (t) = \frac{r(t)}{||r(t)||}$. $x^2 + r(t)^2 = r'^2$ therefore $\sqrt{r'^2 -x^2} = r(t)$. We know that $x(t) = \dot{x}t$, thus $$r(t) = \sqrt{r'^2 - (\dot{x}t)^2}$$
$$\frac{d^2 \mathbf{e}_r}{dt^2} = \frac{d^2}{dt^2} \frac{\sqrt{r'^2 - (\dot{x}t)^2}}{||\sqrt{r'^2 - (\dot{x}t)^2}||}$$
Which is not pretty.
What I really want to say is that $\frac{d^2 \mathbf{e}_r}{dt^2}$ is affected by $x$ and $\ddot{x}= a_x$, but this is delayed by $(t - r/c)$. This, however, seems hand-wavy to me.
 A: The quantity $x$ is not the difference between $\boldsymbol{r}$ and $\boldsymbol{r'}$.  Instead, Prof. Feynman uses $x$ to denote displacement of the charge in a direction perpendicular to $\boldsymbol{e}_{r'}$ (aka a transverse direction), where $\boldsymbol{e}_{r'}$ is defined as the unit vector pointing from the observation point towards the charge (at the retarded time $t-r/c$).  Likewise, $a_x$ is the acceleration of the charge in the transverse direction.  The actual value of the distance $r$ doesn't affect $x$ or $a_x$ (except through determining the retarded time).
So, if the charge happens to be accelerating parallel to $\boldsymbol{e}_{r'}$ (at the retarded time), then $\boldsymbol{e}_{r'}$ doesn't change, so $d^2 \boldsymbol{e}_{r'} / dt^2 = 0$.  Clearly there's also no transverse acceleration in this case, so $a_x=0$, and the formula works.
On the other hand, suppose the charge is accelerating with amplitude $a_x$ transverse to $\boldsymbol{e}_{r'}$ (again at the retarded time).  Now there will be a corresponding acceleration in $\boldsymbol{e}_{r'}$, also in the transverse direction.  How much acceleration? Well, the farther away the charge (the distance $r$), the smaller the acceleration of $\boldsymbol{e}_{r'}$ needed to follow along.  In fact, it's just $a_x/r$, and there's your formula.
