Calculus in the derivation of Euler-Lagrange equation

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SUMMARY

The discussion focuses on the derivation of the Euler-Lagrange equation, specifically addressing the differentiation of the functional S with respect to the parameter α. The key equation discussed is $$\frac{\partial f(Y,Y',x)}{\partial\alpha}=\frac{\partial f}{\partial y}\frac{\partial y}{\partial\alpha}+\frac{\partial f}{\partial y'}\frac{\partial y'}{\partial\alpha}$$, where $$ Y = y(x)+\alphaη(x)$$. The confusion regarding the differentiation stems from the application of the chain rule, clarifying that $$\frac{\partial y}{\partial\alpha}$$ equals η, while the plus sign in the equation arises from the sum of partial derivatives. The participant acknowledges that a Wikipedia article provided the necessary clarification.

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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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