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Suppose that $f(x,y(x),y'(x))$ is s.t $f\in{C^2}$ and $y'(x)\ne{0}.$

I am trying to show that the Euler-Lagrange equation $\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y'}=0$ is equivalent to $\frac{\partial f}{\partial x}-\frac{d}{dx}\big(f-y'\frac{\partial f}{\partial y'}\big) = 0$.

Clearly $f\in{C^2}$ means all the second partial derivatives of $f$ exist and $y'(x)\ne{0}$ is probably necessary for some division by $y'$. I don't know if this is useful but I tried considering when $f-y'\frac{\partial f}{\partial y'}=\frac{\partial f}{\partial y'}.$

Can you give me a possible hint to start me off.

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If we take the total derivative of $f$ w.r.t. $x$ we get that $$\frac{\mathrm{d}f}{\mathrm{d}x}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{\partial f}{\partial y'}\frac{\mathrm{d}y'}{\mathrm{d}x}$$

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  • $\begingroup$ thank you. Also (this could be regarded as a whole separate question, but I expect the answer to be short) what is the notational difference in $\frac{df}{dx}$ and $\frac{\partial f}{\partial x}$ $\endgroup$ – Sam.S May 8 at 13:38
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    $\begingroup$ In the $\mathrm{d}$ case, you are treating it as a function of $x$ only, i.e. you are looking it as $g(x)=f(x,y(x),y'(x))$, while $\partial_x$ looks it as $f(x,y,y')$ and derivates w.r.t. the first argument only. $\endgroup$ – Botond May 8 at 13:54
  • $\begingroup$ thank you for the clarification. $\endgroup$ – Sam.S May 8 at 13:55

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