# showing a different form of the Euler Lagrange Equation provided $f\in{C^2}$ and $y'\ne{0}$

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.

If we take the total derivative of $$f$$ w.r.t. $$x$$ we get that $$\frac{\mathrm{d}f}{\mathrm{d}x}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{\partial f}{\partial y'}\frac{\mathrm{d}y'}{\mathrm{d}x}$$
• 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 $\frac{\partial f}{\partial x}$ – Sam.S May 8 at 13:38
• 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. – Botond May 8 at 13:54