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I'm curious about Euler Lagrange equation. $$d/dx(dF/dy') - dF/dy.$$ $y=y(x)$ why when we partial derivative $dF/dy'$ we don't need to consider $x$ or $y$? like $dF/dx * dx/dy'$

Qmechanic
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hchsmit
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    Because at any given point $x$, the values $y(x)$ and $\dot y(x)$ are independent. The aim of the variational calculus is to end up with such a functional relation between the values and derivatives of $y(x)$. – Lutz Lehmann Sep 25 '22 at 16:17
  • Is ˙(), x also independent? – hchsmit Sep 26 '22 at 00:27
  • Yes, in the linearization step of the action integral variation you just have a function $F(x,y,v)$ of independent variables $x,y,v$. The total derivative you get from the conversion of $\delta v=\delta\dot y$ to $\delta y$ via partial integration, this explicitly deals with changes along the path $y(x)$ and $v(x)=\dot y(x)$. – Lutz Lehmann Sep 26 '22 at 04:45
  • Related: https://math.stackexchange.com/q/580858/11127 and links therein. – Qmechanic Oct 02 '22 at 19:30

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