Simplifying the Euler-Lagrange Equation for Explicitly Independent Functions

[roeb]

Homework Statement

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'} )

Homework Equations

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

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...

 

[CompuChip]

Maybe it helps if you work backwards... what is
<br /> \frac{df}{dx} = \frac{d}{dx} ( y&#039; \frac{\partial f}{\partial y&#039;} )
?