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## Main Question or Discussion Point

I am stuck in trying to understand the derivation of the Euler-Lagrange equation. This mathematical move is really bothering me, I can't figure out why it is true.

[tex]\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}[/tex]

Why is it not:

[tex]\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}+\frac{\partial f}{\partial x}\frac{\partial x}{\partial\alpha}[/tex]

Or better yet:

[tex]\frac{df(y,y';x)}{d\alpha}=\frac{\partial f}{\partial y}\frac{d y}{d\alpha}+\frac{\partial f}{\partial y'}\frac{d y'}{d\alpha}+\frac{\partial f}{\partial x}\frac{d x}{d\alpha}[/tex]

Edit: I think I have found some sound reasoning. That y is a function of x and alpha. So the derivative of f w.r.t. alpha does not involve x. In fact x is held constant for that derivative. That is why dy/d(alpha) and dy'/d(alpha) are partials and not total derivatives.

[tex]\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}[/tex]

Why is it not:

[tex]\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}+\frac{\partial f}{\partial x}\frac{\partial x}{\partial\alpha}[/tex]

Or better yet:

[tex]\frac{df(y,y';x)}{d\alpha}=\frac{\partial f}{\partial y}\frac{d y}{d\alpha}+\frac{\partial f}{\partial y'}\frac{d y'}{d\alpha}+\frac{\partial f}{\partial x}\frac{d x}{d\alpha}[/tex]

Edit: I think I have found some sound reasoning. That y is a function of x and alpha. So the derivative of f w.r.t. alpha does not involve x. In fact x is held constant for that derivative. That is why dy/d(alpha) and dy'/d(alpha) are partials and not total derivatives.

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