What is the derivative of $y$ with respect to $y'$ in general? In the context of variational calculus, we treat $y$ and $y'$ as independent variables and when taking the partial derevitives with respect to each other it gives 0. Is that always the case in general that partial derevitive of $y$ with respect to $y'$ is zero, or it's just for reasons that got to do with variational calculus that we treat $y$ and $y'$ as independent?
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1It is gross notation. For example with something like $L(y,y')$ when you see ${\partial L \over \partial y'}$ it means the partial with respect to the second parameter, not with respect to $y'$ as such. – copper.hat Jun 23 '21 at 14:49
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I see. So y' is like a place holder not really the function y'? After all what does it mean to diffrentiate with respect to a function! Am I correct in thinking that we mean that we should take the partial derivative of the symbol as if it was a variable, but in reality its not a variable its a function? – I0_0I Jun 23 '21 at 14:50
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In this case it is a place holder. Historical. As an aside, it is possible to differentiate with respect to a function if the function is a function of functions if you see what I mean. – copper.hat Jun 23 '21 at 14:52
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So we are taking partial derivatives of the symbol y' not the function y', right? But then we substitute back the meaning of y'? – I0_0I Jun 23 '21 at 14:54
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If not, then what you mean by '' it's a terrible notation "? – I0_0I Jun 23 '21 at 14:58
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When used below the line as in ${\over \partial y'}$ it is purely a place holder. – copper.hat Jun 23 '21 at 14:59
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Does this answer your question? Why is $\frac{\operatorname dy'}{\operatorname dy}$ zero, since $y'$ depends on $y$? – Qmechanic Jun 24 '21 at 16:34
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Yes, it is always zero, $y$ and $y'$ are considered as separate "variables". When you deal with a lagrangian $L$ you usually consider it with variables $(x,s,p)$ with $x \in \mathbb{R}^n$, $s \in \mathbb{R}$ and $p \in \mathbb{R}^n$ (I am talking about the scalar case here, I don't want to complicate things with vector valued functions). What you have is that $s$ represents the value of your function $u: \mathbb{R}^n \to \mathbb{R}$ while $p$ represent the partial derivatives of $u$ with respect to $x_1 , \dots, x_n$.
Notice that here we don't use $y$ and $y'$ because it would generate great confusion, it's not a wise notation at all. The best exercise to help you understand these things is to derive the Euler-Lagrange equation by yourself.
tommy1996q
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