Functional Derivatives

Euler-Langrange looking thing

This calculation looks wurry wurry emportunt, because we have a Lagrangian looking function g(y, f(x), f'(x)) that we are integrating in the functional, and we want to find its functional derivative.

Why is that important? I think it has to do with a way to describe the motion of an object, given its equation of state.

p17

 Using their definition of functional derivative gets tricky because it involves delta functions.

p12

 source: QFT for the gifted noob.

Here's my attempt:

I vary f(x) by epsilon times a delta (y-x) term. By analogy I vary the f '(x) term by epsilon times delta '(y - x).
Next, I Taylor expand the function g to get first order in epsilon terms.
In order to evaluate the delta '(y-x) term, I integrate by parts.
 
This gets me close to the answer. I'm skeptical of the math involved, but physics explanations usually are dubious about doing things that make math sense. Problem is I get an extra term.

I'm getting a partial derivative term with respect to f'(x) multiplied by a delta function at y=x, which if that equals zero somehow, then my result equals the desired Euler-Lagrange term for the partial derivative.

Same thing for the g(y, f, f', f'') case; there's an extra term dg/df'' *delta ' (x' - x) in my calculation, that were it equal to zero for some reason, I would get the prescribed dosage of partial derivative terms in the result.