Euler-Lagrange equation: Differentiation with respect to x

Question

I got stuck in my lecture notes after a supposed differentiation of the Euler-Lagrange equation: $$\dfrac{\partial f}{\partial y}-\dfrac{d}{dx} \left( \dfrac{\partial f}{\partial y'}\right) = 0$$

Differentiate with respect to $x$ to obtain:

$$\dfrac{\partial^2 f}{\partial y'^2}y''+\dfrac{\partial^2 f}{\partial y'\partial y}y'+\left( \dfrac{\partial^2 f}{\partial y' \partial x} - \dfrac{\partial f}{\partial y} \right) = 0$$

It's been a while since I've been baffled by differentiation. I am guessing I am not familiar with the notation enough.

Any hints are welcome, I (hopefully) only need a nudge in the right direction.

Thanks!

Answer 1

$$ \dfrac{\partial f}{\partial y} -\dfrac{d}{dx}\dfrac{\partial f}{\partial y'} = 0\tag{1} $$ since $$ \dfrac{d}{dx} = \dfrac{\partial}{\partial x} + y'\dfrac{\partial}{\partial y} + y''\dfrac{\partial}{\partial y'} $$ thus Eq. (1) becomes $$ \dfrac{\partial f}{\partial y} - \left[\dfrac{\partial}{\partial x} +y'\dfrac{\partial}{\partial y} + y''\dfrac{\partial}{\partial y'}\right]\dfrac{\partial f}{\partial y'} =\dfrac{\partial f}{\partial y} -\dfrac{\partial^2 f}{\partial x\partial y'} - y'\dfrac{\partial^2f}{\partial y \partial y'} - y''\dfrac{\partial^2f}{\partial y'^2} $$

Answer 2

The thing is that $f = f(x,y, y')$ and $y=y(x)$. When using partial derivatives you must think of $y$ and $y'$ as independent variables (no chain rule), but when using total derivatives use the chain rule. So, by the chain rule:

$$ \frac{\partial f}{\partial y} - \frac{d}{dx} \frac{\partial f}{\partial y'} =\frac{\partial f}{\partial y} - \frac{\partial^2 f}{\partial y'\partial x} - \frac{\partial^2 f}{\partial y'\partial y}y' -\frac{\partial^2 f}{\partial y'^2} y''$$

Since

$$\frac{d}{dx} \frac{\partial f}{\partial y'} = \frac{\partial }{\partial x}\frac{\partial f}{\partial y'} - y'\frac{\partial }{\partial y}\frac{\partial f}{\partial y'} -y''\frac{\partial }{\partial y'}\frac{\partial f}{\partial y'} y''$$