Undergrad Calculus in the derivation of Euler-Lagrange equation

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In the derivation of the Euler-Lagrange equation, the differentiation of S with respect to α involves the chain rule, leading to the expression that includes both partial derivatives of f with respect to y and y'. The confusion arises from the notation, where the correct differentiation with respect to α is indeed related to y and not Y, as Y incorporates the perturbation η. The plus sign in the equation is a result of applying the chain rule to both components of the function f. The discussion highlights that understanding this derivation is crucial for grasping the integrand when α equals zero. Overall, the clarification of these concepts is essential for a deeper understanding of the Euler-Lagrange equation.
BearY
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In the derivation of Euler-Lagrange equation, when differentiating S with respect to α, there is a step:
$$\frac{\partial f(Y,Y',x)}{\partial\alpha}=\frac{\partial f}{\partial y}\frac{\partial y}{\partial\alpha}+\frac{\partial f}{\partial y'}\frac{\partial y'}{\partial\alpha}$$
Where $$ Y = y(x)+\alphaη(x)$$

My puny math knowledge can't tell me 2 things:
1.why is it ##\frac{\partial y}{\partial\alpha}## instead of ##\frac{\partial Y}{\partial\alpha}##? Isn't the second one equal to η? Why is the first one equal to η? Did I skip something?

2. Where does the plus sign come from? I learned partial derivative before but I cannot recall anything like this. I have a feeling this is the result of forgetting something completely:oops:

Edit: NM the second one I was stupid It's just chain rule :oops:
 
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Wikipedia does a good write-up. They use the letter ## g ## instead of ## Y ##, but comparing the two, you are correct that it should be a capital ## Y ## and ## Y' ## in those terms.
 
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Charles Link said:
Wikipedia does a good write-up. They use the letter ## g ## instead of ## Y ##, but comparing the two, you are correct that it should be a capital ## Y ## and ## Y' ## in those terms.
That Wikipedia page really helped, thanks. I see that ##\frac{\partial S}{\partial\alpha}## when ##\alpha = 0## is the integrand we are looking for. My book didn't bother explaining that directly.
 
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