Feynman Lecture Principle of Least Action: Glossed over Taylor expansion? His initial one dimensional derivation of Newton's Second Law using the Principle of Least Action, I believe is fairly concise and easy to read. However, I did get hung up on his use of the Taylor series expansion, and I noticed this quote from his lecture:

I have written $V′$ for the derivative of $V$ with respect to $x$ in order to save writing. 

This is how I come about the same expansion for the potential function, $V$:
$x=\underline{x}+\eta$
Where x is the approximated path taken by a classical particle, $\underline{x}$ is the actual path, and $\eta$ is the error (which is zero at the start and finish point). Expanding the first two terms of the taylor series for V,
$V(\eta)_{centered\ on\ \underline{x}} \approx V(\underline{x}) + V'(\underline{x})(\eta - \underline{x})$
$V(\eta + \underline{x}) \approx V(\underline{x}) + V'(\underline{x})\eta$
Which is, no surprise, the same thing noted in the lecture. I won't bother with following the math through to the end result (it's simple), but I will note that the derivative of the potential function, V, evaluated at $\underline{x}$ shows up in the final result:
$m\ddot{\underline{x}} = -V'(\underline{x})$
And I have been scratching my head at this for many days (far too many days). The original potential function should have been with argument $\eta$, centered on the actual path, $\underline{x}$. This is counter to my notion that the potential functional should be dependent only on a position in real space. This implies greatly that,
$m\ddot{\underline{x}} = -\frac{dV}{d\eta}\bigg{|}_{\eta=\underline{x}}$
Revealing a force function as follows,
$F = -\frac{dV}{d\eta}\bigg{|}_{\eta=\underline{x}}$
Can someone explain what this means? Why would the force function be proportional to the derivative of the potential function with respect to the error in the path compared to the actual path taken?
 A: This is one of those places where almost-universally helpful notation can trip you up.
$V$ is a function.  It eats a single real number and then spits out another real number.  As a result, you can differentiate it with respect to one thing and one thing only - its argument.  The result is another function, which we give the related name $V'$.
$V(\underline x + \eta)$ is a number - specifically, the number which $V$ spits out when you feed it the number $\underline x + \eta$.  If $V$ is sufficiently well-behaved and $\eta$ is sufficiently small, Taylor's theorem tells us that $V(\underline x + \eta)$ can be approximated as follows:
$$V(\underline x + \eta) \approx V(\underline x) + V'(\underline x)\cdot  \eta $$
In words,

The function $V$ evaluated at the number $\underline x + \eta$ is approximately equal to the function $V$ evaluated at the number $\underline x$, plus $\eta$ times the function $V'$ evaluated at the number $\underline x$.

Hopefully this doesn't come across as overly mathematical and pedantic - the point I'm trying to convey is that you are confusing yourself by writing $\left.\frac{dV}{d\eta}\right|_{\eta = \underline x}$.  It doesn't make sense to talk about differentiating $V$ with respect to anything other than its argument.  In that expression, $\eta$ is essentially a dummy variable - it means exactly the same thing as $\left.\frac{dV}{dy}\right|_{y=\underline x}$ or $\left.\frac{dV}{d\star}\right|_{\star = \underline x}$.
