Questions regarding the derivation of Euler-Lagrange Equation from Taylor's Classical Mechanics I'm self-studying classical mechanics from Taylor's book. I saw his derivation of the Euler-Lagrange Equation and I'm confused about something, he created a 'wrong' function $$Y(x) = y(x)+\eta(x)\tag{6.6}$$
Where $y(x)$ is the minimal path and $\eta(x)$ is some arbitrary function.
He then argues that since $y(x)$ is the minimum any other function, no matter how close to $y(x)$ must be greater than $y(x)$.
So using this fact, he parameterizes the action integral $S$ by $\alpha$. That is,
$$S(\alpha)= f(y+\alpha \eta, y' + \alpha \eta, x).$$
My confusion is that, later in the chapter he says this the Euler-Lagrange Equation finds the stationary point not the minimum but the derivation used the fact that for all $\alpha$ other than zero, $y<Y$. How does that work?
Secondly, in order for $S(\alpha)$ to be stationary he puts,
$$\frac{dS}{d\alpha} = \int \frac{\partial f(y+\alpha \eta, y'+\alpha \eta ', x)}{\partial \alpha}.$$
Here since $\alpha$ appears in two arguments of $f$, the derivative will be (using chain rule):
$$\frac{\partial S}{\partial \alpha} = \eta \frac{\partial f}{\partial y} + \eta' \frac{\partial f}{\partial y'}$$
I don't understand this step. How do you know you'll have two terms? And also where did that addition sign came from? Also shouldn't it be the partial with respect to $Y$ (the wrong function) and not $y(x)$?
The book I'm referring to is Classical mechanics by John R. Taylor
 A: 
How does that work?

A minimum is a stationary point (a point where the first derivative is zero). If you want to see whether it is a maximum or a minimum (or a saddle point) look at the second derivative.

$$\frac{\partial S}{\partial \alpha} = \eta \frac{\partial f}{\partial y} + \eta' \frac{\partial f}{\partial y'}$$
I don't understand this step. How do you know you'll have two terms? And also where did that addition sign came from?

This is just basic multivariable derivative calculus.
You have a function of three arguements (call them $a$, $b$, and $c$).
$$
f(a, b, c)
$$
By definition $\frac{\partial f}{\partial a}$ means the derivative wrt to the first argument with the others held fixed. This tells me how much $f$ changes if I just change the first argument by an amount $\delta a$:
$$
(\delta f)_{a\; only} = \frac{\partial f}{\partial a}\delta a
$$
Similarly, if I just change the second argument by $\delta b$ then $f$ change by:
$$
(\delta f)_{b\; only} = \frac{\partial f}{\partial b}\delta b
$$
If I change both $a$ to $a+\delta a$ and $b$ to $b+\delta b$ the total amount $f$ changes is:
$$
(\delta f)_{a\;and\;b} = \frac{\partial f}{\partial a}\delta a + \frac{\partial f}{\partial b}\delta b\;.
$$
In your case, because $\alpha$ is a parameter in both the first and second argument, when you change alpha you change both the first and second argument and you need to account for this by summing.
A: It is not clear to me what is/are the confusions here, but I will try to clear the concerns I see...
The action being stationary is another way of saying "we evaluate the action at the path for which it (i.e. the action) is minimum".
The action is a functional (its arguments are functions themselves, functions of $x$ that is). So its variation is not given by ordinary derivatives, but by something we call functional derivatives. Notation-wise, this means that the differential sign $d$ is replaced by a $\delta$ (i.e. $\frac{d}{dx}\rightarrow\frac{\delta}{\delta y}$, where $y$ must now be a function of some variable! Everything works pretty much the same (but I am neglecting a lot of details when I am saying that, so hopefully no one comes at me for saying that).
In the variation of the action, the author assumes that he/she has a dependence of the action only on a function $y(x)$ and its first derivative! This might not be explained explicitly, but I think it is assumed... So, the latter function and its derivative count as two independent functions and in the same way one would Taylor expand a function of two variables by writting it in terms of the partial derivative wrt the first variable times the differential associated to the first variable, plus the partial derivative wrt the second variable times the differential associated to the second variable, one can also Taylor expand functionals... This is what is going on here (I would substitute the partial signs for greek deltas to indicate that I am differentiating wrt functions if I were you, but this is not the point).
Last, but not least, no, it shouldn't be a partial with respect to $Y(x)$, as you are Taylor expanding around $y(x)$. In ordinary Taylor expansions of functions around some variable $x$, we know that
$$f(x+a)=f(x)+a\frac{df}{dx}\Big|_{a=0}$$
and the same goes here. The role of $Y(x)$ is played by $x+a$, the role of $y(x)$ by $x$ and the role of $a\eta(x)$ by $a$.
I hope things make more sense now. If I didn't get any of your questions, please comment.
