This is mostly a re-hash of this thread, but it did not receive an adequate answer. In the derivation that I am reading, there is one step that is not justified. Perhaps obvious, but it is not clear to me. Here is the build-up:
Suppose we have the following (one-dimensional) integral that we want to minimize by finding a suitable function $Y(x)$: $\int_{x_1}^{x_2} f(Y(x),Y'(x),x) dx$. Assume we already know the function that minimizes this to be $y(x)$. Define $Y(x,\alpha) = y(x)+\alpha \eta(x)$ and require $\eta(x)$ to be a differentiable function with $\eta(x_1) = \eta(x_2) = 0$ so that this is a suitable path between the points of interests, and assume $\alpha$ is real. From this, define
$$S(\alpha) = \int_{x_1}^{x_2} f(Y(x,\alpha),Y'(x,\alpha),x) dx$$
It is clear that the minimum path is simply $Y(x,0)$, which implies that $\frac{\partial S}{\partial \alpha}\Big|_{\alpha = 0} = 0$. Now, here is where my confusion is:
With no dependence on $\alpha$ for the bounds, we know that
$$\frac{\partial S}{\partial \alpha}\Big|_{\alpha = 0} = \Big[\int_{x_1}^{x_2} \frac{\partial f}{\partial \alpha} dx \Big]_{\alpha = 0} = \Big[\int_{x_1}^{x_2} \Big(\eta(x)\frac{\partial f}{\partial Y}+\eta'(x)\frac{\partial f}{\partial Y'}\Big) dx\Big]_{\alpha = 0} = 0$$
and my author, without justification, immediately makes the jump to
$$\Big[\int_{x_1}^{x_2} \Big(\eta(x)\frac{\partial f}{\partial Y}+\eta'(x)\frac{\partial f}{\partial Y'}\Big) dx\Big]_{\alpha = 0} = \int_{x_1}^{x_2} \Big(\eta(x)\frac{\partial f}{\partial y}+\eta'(x)\frac{\partial f}{\partial y'}\Big) dx$$
Perhaps in clearer form, the following step is taken:
$$ \frac{\partial f(y+\alpha \eta,y'+\alpha \eta',x)}{\partial \alpha} = \eta \frac{\partial f}{\partial y} + \eta'\frac{\partial f}{\partial y'}$$
Can someone provide justification for this step? My intuition tells me that
$$\frac{\partial f}{\partial Y(x,\alpha)}\Big|_{\alpha = 0} = \frac{\partial f}{\partial Y(x,0)} = \frac{\partial f}{\partial y(x)} $$
but some essential part of that just feels wrong. It reminds me of the silly mistake that a careless calculus student can make, saying that
$$\frac{\partial f(x,y)}{\partial x}\Big|_{(x,y) = (x_o,y_o)} = \frac{\partial f(x_o,y_o)}{\partial x}$$
Which is clearly incorrect by noting that $f(x_o,y_o)$ is simply some constant. In the above case I think the issue is much subtler.
