Derivation of Euler-Lagrange equation Here is  a simple (probably trivial) step in the derivation of the Euler-Lagrange equation.
    If we denote $Y(x) = y(x) + \epsilon \eta(x) $, I want to know why is 
$\dfrac{\partial f(Y,x)}{\partial Y} \Big\vert_{\epsilon = 0} = \dfrac{\partial f(y,x)}{\partial y} $
Could someone justify the  steps involved in justifying this ? I am sure am missing something elementary.
EDIT: Could someone comment on correctness of the 'proof ' ?
    $\dfrac{\partial f(Y,x)}{\partial Y} \Big\vert_{\epsilon = 0} = \lim_{H\rightarrow 0} \dfrac{f(y + \epsilon \eta  + H,x) -f(y + \epsilon \eta,x)}{H} \Big\vert_{\epsilon = 0} = \lim_{H\rightarrow 0} \dfrac{f(y + H,x) -f(y,x)}{H} = \dfrac{\partial f(y,x)}{\partial y} $
EDIT 2:
The answers provided still leave me confused. Here is another attempt at a "proof".
$\dfrac{\partial f(Y,x)}{\partial Y} \Big\vert_{\epsilon = 0} = \lim_{\epsilon\rightarrow 0} \dfrac{\partial f(Y,x)}{\partial Y} = \lim_{\epsilon \rightarrow 0} \left(\lim_{H\rightarrow 0} \dfrac{f(y + \epsilon \eta  + H,x) -f(y + \epsilon \eta,x)}{H} \right)$
Now if I could interchange the limits, then it would make sense that I get
  $\lim_{H\rightarrow 0}  \dfrac{f(y + H,x) -f(y,x)}{H} = \dfrac{\partial f(y,x)}{\partial y} $.
So is the interchange of limits allowed. What hypothesis must be satisfied for that to happen ?
 A: This Confusion with Euler-Lagrange Derivation was basically what convinced me that it was all a muddle with symbols and notation.  Just for the benefit of future readers, I write it down in my own words.
So, the notation $ \dfrac{d f}{dx}\Big|_{x = x^*}$ basically means derivative of the function $f(\cdot)$ with respect to its argument, and then evaluation of the result when the argument is set to $x^*$.
Exactly the same reasoning is involved in the question, for we take  the partial derivative  with respect to $Y$ and the result is evaluated at $\epsilon =0$.
But evaluation of the result at $\epsilon =0$ is equivalent to 
$ \dfrac{\partial f(Y, x)}{\partial Y}\Big|_{Y = y}$ which looks silly, but it is blindingly obvious to me now that  
$ \dfrac{\partial f(Y, x)}{\partial Y}\Big|_{\epsilon = 0} = \dfrac{\partial f(Y, x)}{\partial Y}\Big|_{Y = y} = \dfrac{\partial f(y, x)}{\partial y}$
A: The formula for $\frac{\partial f(Y,x)}{\partial Y}|_{\epsilon=0}$ is not correct: you should write
$$\frac{\partial f(Y,x)}{\partial Y}|_{\epsilon=0}:=
\lim_{\epsilon\rightarrow 0}\frac{
f(y+\epsilon\eta,x)-f(y,x)}{\epsilon}=
\lim_{\epsilon\rightarrow 0}\frac{
\langle \nabla f(y,x),h_\epsilon\rangle+O(\|h\|^2)}{\epsilon},
$$
where we suppose that $f$ is differentiable at $(y(x)+\epsilon\eta(x),x)$ for all $x$, denoting by $h_\epsilon$ the increment
$$h_\epsilon=( \epsilon\eta(x), 0)$$
and by $\nabla f(y,x)=\left(\frac{\partial f}{\partial y},\frac{\partial f}{\partial x}\right)$ the gradient of $f$ at $(y(x),x)$.
Then 
$$\frac{\partial f(Y,x)}{\partial Y}|_{\epsilon=0}=
\lim_{\epsilon\rightarrow 0}\frac{\frac{\partial f}{\partial y}\epsilon\eta(x)
+O(\epsilon^2)}{\epsilon}=\frac{\partial f}{\partial y}(y(x),x)\eta(x),
$$
as claimed (for all small variations $\eta(x)$).
What we are missing here are the boundary conditions on the increment $\eta$ of $y$ in order to talk about a "variational problem" and the explicit form of the functional $f$.
