Calculus of variations and least action from Feynman lectures. In Feynman's lecture on least action, he mentions that a perturbation at an extremum of the action functional would cause a zero first order change in the action. Intuitively this makes sense because I find it analogous to the Taylor expansion of a function - at extrema, $f^\prime = 0$ this the first order term disappears. My difficulty is with the fact that the action is a functional, and thus I don’t really understand this rigorously.
http://www.feynmanlectures.caltech.edu/II_19.html
Furthermore, the Wikipedia derivation of the Euler Lagrange equation seemed to make sense to me. However, I am unable to connect that derivation to statement made by Feynman.
 A: The proof is very easy if you know the multivariable chain rule (for the Calculus of Variations, you need to know the infinite-dimensional version... which is still the same) and the single variable version of this theorem. The statement

he mentions that a perturbation at an extremum of the action functional would cause a zero first order change in the action

made by Feynman would be phrased in the following manner by a Math person:

Theorem:
Let $V$ be a real Banach space (a complete normed vector space), and let $f:V \to \Bbb{R}$ be a differentiable function. If $x_0$ is a local extremum point for $f$, then $df_{x_0} = 0$.

Note that the differential $df_{x_0}$ is itself by definition a continuous linear transformation from $V$ into $\Bbb{R}$. So, what we have to show is that for every "perturbation" $\xi \in V$, we have $df_{x_0}(\xi) = 0$. So, to prove this, pick any $\xi \in V$. Now, define the curve $\gamma : \Bbb{R} \to V$ by $\gamma(t) = x_0 + t\xi$. Now, consider the composite mapping $f \circ \gamma: \Bbb{R} \to \Bbb{R}$. It is clear that $\gamma$ is differentiable, and by assumption, $f$ is also differentiable. Hence, by the chain rule, $f \circ \gamma$ is differentiable at the origin.
Next, note that since $x_0$ is a local extremum of $f$, it follows that $0$ is a local extremum for $f \circ \gamma$. Since $f\circ \gamma$ is a simple mapping from $\Bbb{R} \to \Bbb{R}$, we can use the simple single-variable calculus version of the theorem to conclude that $(f \circ \gamma)'(0) = 0$. Hence,
\begin{align}
0 &= (f \circ \gamma)'(0) \\
&= df_{\gamma(0)}\big( \gamma'(0)\big) \tag{chain rule} \\
&= df_{x_0}(\xi) \tag{by definition of $\gamma$}
\end{align}
This is precisely what we wanted to prove! Since $\xi$ was arbitrary, this completes the proof that $df_{x_0} = 0$.

The theorem proven above is very general. In the particular context of the principle of "least" action, here's how we interpret the above theorem. The Banach space $V$ is usually a set of curves in the configuration space of the system. We interpret $f$ to be the action (I realize the typical notation is $S$ or $J$), and what I called $x_0$ is going to be a " local extremal curve" for the action $f$.
By the way, the Euler Lagrange equations have absolutely nothing to do with the actual statement Feynman made. The statement made by Feynman translated word-for-word into more pedantic and technical mathematical jargon is exactly what I have written in the Theorem above.
What the Euler Lagrange equations tell you is that if the action $S: V \to \Bbb{R}$ is of the specific form
\begin{align}
S(\varphi) := \int_a^b L(t, \varphi(t), \varphi'(t)) \, dt
\end{align}
where $L: \Bbb{R} \times \Bbb{R} \times \Bbb{R} \to \Bbb{R}$ is a smooth function,
then the condition $dS_{\varphi} = 0$ is equivalent to the Euler Lagrange equations:
\begin{align}
\dfrac{d}{dt} \bigg( \big(\partial_3L\big) \big(t, \varphi(t), \varphi'(t) \big) \bigg) - \big(\partial_2 L \big)(t, \varphi(t), \varphi'(t)) = 0
\end{align}
Where $\partial_iL$ means the $i^{th}$ partial derivative of $L$.
