I am studying calculus of variations, and I see functions of this sort quite a bit: $$F(x,y,\dot{y})$$ Why is $\dot{y}$ mentioned in these functions? Is it incomplete to simply state that $F(x,y,\frac{dy}{dx})$ is equivalent to some $G(x,y)$?
Also, let's say I introduce a variable $\bar{y}$, such that $$\bar{y}(x) = y(x) + \epsilon \eta (x)$$ for some $\epsilon \in \mathbb{R}$ and $\eta$ is some arbitrary function of $x$ And decide to differentiate $F(x,\bar{y},\frac{d\bar{y}}{dx})$ wrt $\epsilon$. I expected something like this from the chain rule: $$\frac{dF}{d\epsilon} = \frac{dF}{dx} \frac{dx}{d\epsilon} + \frac{dF}{d\bar{y}} \frac{d\bar{y}}{dx} \frac{dx}{d\epsilon} + \frac{dF}{d\bar{y}'} \frac{d\bar{y}'}{dx} \frac{dx}{d\epsilon}$$ However, from this video: https://www.youtube.com/watch?v=sFqp2lCEvwM&list=PLdgVBOaXkb9CD8igcUr9Fmn5WXLpE8ZE_&index=2 at 4:46, you can see that $$\frac{dF}{d\epsilon} = \frac{dF}{d\bar{y}} \frac{d\bar{y}}{d\epsilon} + \frac{dF}{d\bar{y}'} \frac{d\bar{y}'}{d\epsilon}$$.
Of course, in the video they are deriving the Euler-Lagrange equation, so $x$ does not depend on $\epsilon$ and gets cancelled, but the other terms survive.
What am I missing here?
Why is $\dot{y}$ is mentioned in these functions? Is it incomplete to simply state that $F\left(x,y,\frac{dy}{dx} \right)$ is equivalent to some $G(x,y)$?
$F(x,y,\dot{y})$ is a function of $3$ variables. These variables are independent. We use this function as a functional that depends on variable $x$, function $y$ and its first derivative. Notation can be a little bit misleading (we just denote the third variable as $\dot{y}$). So it is not equivalent to some function $G(x,y)$, because we're considering arguments $y$ and $\dot{y}$ as independent (even though eventually third argument $\dot{y}$ appears to be first derivative of $y$).
Regarding your second questing, as already has been pointed in comments, $x$ does not depend on $\varepsilon$, so applying chain rule you get $$ \frac{dF}{d\epsilon} = \frac{dF}{d\bar{y}} \frac{d\bar{y}}{d\epsilon} + \frac{dF}{d\bar{y}'} \frac{d\bar{y}'}{d\epsilon}. $$