Calculus of Variation: Chain Rule and Formulation Proof

[Gavroy]

I have a question about calculus of variation.

does anybody here know a proof for the chain rule:
[tex]\delta S= \frac{dS}{dx} \delta x[/tex]

and for the formulation:

[tex]\delta S= p \delta x[/tex]
=> [tex]\frac{dS}{dx}= p [/tex]

it would be totally sufficient, if anyone here knows(e.g. a weblink) where one could see this proof.

 

[EnumaElish]

I find your notation confusing. Are you asking the conditions under which dy/dx = [partial]y/[partial]x ?

 

[Gavroy]

No, it is about calculus of variation and the delta should be the gateaux derivative

 

[dextercioby]

Ok, if the delta S in the LHS is the [itex] \mbox{G}\hat{\mbox{a}}\mbox{teaux} [/itex] derivative of a functional S, then what are the things in the RHS ?

 

[Gavroy]

sorry that my notation confused you.

if I say [tex]S(x)=p \cdot x[/tex]
then the variation says:
[tex]\delta S(x)=p \frac{d(x+\epsilon h)}{d \epsilon}|_{\varepsilon=0}[/tex]
or in the other notation:
[tex]\delta S(x)=p \cdot \delta x[/tex]

and now I wanted to see the proof of the chain rule:

[tex]\delta S(x)=\frac{\partial S}{\partial x} \cdot \delta x[/tex]

and that:[tex]\frac{\partial S}{\partial x}=p[/tex]

but I guess both proofs go hand in hand