Understanding step in the derivation of Euler-Lagrange There is a step in what seems to be the standard derivation of the Euler-Lagrange equation that I do not see how can be justified (it is the first derivation in the Wikipedia article on the equation).
We have that $g(\epsilon) = f(t) + \epsilon \lambda(t)$
We are computing the total derivative $\frac{\partial F}{\partial\epsilon}(t,g(\epsilon),\dot{g}(\epsilon))$ and get that this is equal to $\frac{\partial F}{\partial g}\lambda(t) + \frac{\partial F}{\partial \dot{g}}\dot{\lambda}(t)$. The thing that does not make sense to me is that this is supposed to equal $\frac{\partial F}{\partial f}\lambda(t) + \frac{\partial F}{\partial \dot{f}}\dot{\lambda}(t)$ at $\epsilon = 0$.
To make some sense of it, I tried using the definition of the derivative on $\frac{\partial F}{\partial g}\lambda(t)$ to see if I can make it equal $\frac{\partial F}{\partial f}\lambda(t)$. However, what I do get is:
$$\lim_{\Delta \epsilon \to 0}\frac{F(t,f(t)+(\epsilon+\Delta \epsilon) \lambda(t),\dot{g}) - F(t,f(t)+ \epsilon \lambda(t),\dot{g})}{\Delta \epsilon }$$
$$=\lim_{\Delta \epsilon \to 0}\frac{F(t,f(t)+(\epsilon+\Delta \epsilon) \lambda(t),\dot{g}) - F(t,f(t)+ \epsilon \lambda(t),\dot{g})}{\Delta g } \frac{\Delta g}{\Delta\epsilon}$$
Here we have that $\Delta g = \Delta \epsilon \lambda(t)$. This gives me:
$$=\lim_{\Delta \epsilon \to 0}\frac{F(t,f(t)+(\epsilon+\Delta \epsilon) \lambda(t),\dot{g}) - F(t,f(t)+ \epsilon \lambda(t),\dot{g})}{\Delta g } \lambda(t)$$
Now I am just stuck. I cannot convince myself that $\Delta g = \Delta f$. I mean, the derivation is with respect to $\epsilon$, not t. So, why does this lead to the Euler-Lagrange equation?

Addendum
I think I may have to clarify my problem:
$\frac{\partial F}{\partial f} \lambda$ looks to me to be equal to:
$$\lim_{\Delta f \to 0}\frac{F(t,f(t)+\Delta f(t),\dot{g}) - F(t,f(t),\dot{g})}{\Delta f(t) } \lambda(t)$$
However, my computations above imply that the solution is
$$\lim_{\Delta \epsilon \to 0}\frac{F(t,f(t)+(\epsilon+\Delta \epsilon) \lambda(t),\dot{g}) - F(t,f(t)+ \epsilon \lambda(t),\dot{g})}{\Delta g } \lambda(t) = \lim_{\Delta \epsilon \to 0}\frac{F(t,f(t)+\Delta \epsilon \lambda(t),\dot{g}) - F(t,f(t),\dot{g})}{\Delta \epsilon \lambda(t) } \lambda(t) $$
Clearly, my reasoning and computations are wrong. However, I don't see how I can get the solution above by computing $\frac{dF}{d\epsilon}$.
 A: It's just the chain rule. You are given a function $F(\cdot,\cdot,\cdot):{\mathbb R}^3\to{\mathbb R}$ of three variables, say $t$, $p$, $q$, and then are studying the expression $$F\bigl(t,f(t),\dot f(t)\bigr)$$
for various known or unknown "test functions" $t\mapsto f(t)$. In particular you form the expression
$$\psi(\epsilon):=F\bigl(t,f(t)+\epsilon g(t),\dot f(t)+\epsilon\dot g(t)\bigr)\ .$$
Then for any fixed $t$ you have
$$\psi'(\epsilon)=F_{.2}\bigl(t,f(t)+\epsilon g(t),\dot f(t)+\epsilon\dot g(t)\bigr) \>g(t)+F_{.3}\bigl(t,f(t)+\epsilon g(t),\dot f(t)+\epsilon\dot g(t)\bigr)\>\dot g(t)\ .\tag{1}$$
Note that you do not "differentiate $F$ with respect to $f$" here. It is an abuse of notation to write ${\partial F\over\partial f}$, but we have learnt to live with it. $F$ has known partial derivatives $F_{.i}$ at each point $(t,p,q)\in{\mathbb R}^3$ before any "test function" $f$ is inserted.
Put $\epsilon:=0$ in $(1)$ and obtain
$$F_{.2}\bigl(t,f(t),\dot f(t)\bigr) \>g(t)+F_{.3}\bigl(t,f(t),\dot f(t)\bigr)\>\dot g(t)\ .$$
