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assuming that $y$ is a function of $x$, would this work:

$$\frac{\partial F}{\partial \dot{y}}\dot{x} = \frac{\partial F}{\partial\left(\dot{y}/\dot{x}\right)} = \frac{\partial F}{\partial y’}, y’ = \frac{\dot{y}}{\dot{x}} = \frac{dy}{dx}$$

Here, $y$ and $x$ are parametrized by $t$, I.e., $\dot{y} = dy/dt$, and $F$ is a function of $x$, $\dot{x}$, $y$, and $\dot{y}$.

I saw this from A First Course in The Calculus of Variations by Mark Kot.

Superman
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  • What do you mean by $\dot{x}$ and $\dot{y}$? The final term, $\frac{dy}{dx}$ doesn't appear to be a function of F. What happened to F? Do you mean that y= F(x)? – user247327 Jan 05 '21 at 18:56
  • No, F is a function of y and x. And I am using t to parametrize x and y. – Superman Jan 05 '21 at 18:57
  • Hi, I have updated the query. – Superman Jan 05 '21 at 23:11
  • Uh, where in Kot's book did you read this? It looks like you're trying to rewrite the Euler-Lagrange equation, but have erred somewhere. – Alex Nelson Jan 05 '21 at 23:45
  • I think that I should re-formulate my query regarding the book in this case then. Sorry about it! Will get to it later. – Superman Jan 05 '21 at 23:53
  • Hi, I realized that in Kot’s book page 201, he made a typo; he put the wrong independent variable for the chain rule. It should have been $\dot{y}/\dot{x}$. – Superman Jan 06 '21 at 06:51

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