# Euler-Lagrange Simplification

1. Jan 25, 2009

### roeb

1. The problem statement, all variables and given/known data

If the integrand f(y, y', x) does not depend explicitly on x, that is, f = f(y, y') then
$$\frac{df}{dx} = \frac{\partial f}{\partial y}y' + \frac{ \partial f } {\partial y' } y''$$

Use the Euler-Lagrange equation to replace $$\partial f / \partial y$$ on the right and hence show that $$\frac{df}{dx} = \frac{d}{dx} ( y' \frac{\partial f}{\partial y'} )$$
2. Relevant equations

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

3. The attempt at a solution

By substituting in for df/df, I get an extra term that I can't seem to make go away.

$$\frac{df}{dx} = \frac{d}{dx} y' \frac{ \partial f }{\partial y'} + \frac{\partial f}{\partial y'} y''$$

I can't seem to get rid of that extra term, it seems like it should be straight forward but...

2. Jan 26, 2009

### CompuChip

Maybe it helps if you work backwards... what is
$$\frac{df}{dx} = \frac{d}{dx} ( y' \frac{\partial f}{\partial y'} )$$
?