Derivations of Euler equations

[zwoodrow]

Every text book i find breezes over the following point:

\delta\partial (x) =\partial \delta (x)

where delta is just the variation. Someone asked me why thats true and i guessed the only thing i could say was that delta is an operation not a variable so this is more like an algebraric statment of the commutivity of delta and taking a derivative. Is this right or is there a more clear explanation.

[CompuChip]

I always view it as follows.
What we do, is changing some function f(x) by an infinitesimal amount, i.e. replace
f(x) \to f(x) + \delta f(x) \qquad\qquad(*)
(and ignore higher order variations).

So in this notation, where delta indicates the infinitesimal change, it should also be true that
\frac{\partial f(x)}{\partial x} \to \frac{\partial f(x)}{\partial x} + \delta \left(\frac{\partial f(x)}{\partial x} \right)
(which is just statement (*) again for another function g(x) = df(x)/dx).

Now if you differentiate (*) you get an equation from which it follows immediately that
\delta \left(\frac{\partial f(x)}{\partial x} \right) = \frac{\partial(\delta f(x))}{\partial x}

[zwoodrow]

Is there a way to argue that (using the chain rule)
\partial(\deltaf(x) ) = f(x)\partial\delta + \delta\partialf(x)
and then argue that\partial\delta is a second order differential that can be tossed out as approx 0 giving the result

\partial(\deltaf(x) ) = f(x)\partial\delta \rightarrow 0 + \delta\partialf(x)

[CompuChip]

Not if you want to be rigorous.
Because \delta in itself doesn't mean anything, just like \partial doesn't mean anything.

The object you are looking at is \delta f(x).
If you want to be precise, you can use a first order Taylor expansion
f(x) \to F(x) = f(a) + f'(a) (x - a) + \mathcal{O}((x - a)^2)
where you call \delta = f'(a) (x - a) and neglect the quadratic terms.