The rest of the article leads me to believe that it's just like a normal derivative, except that it's being taken with respect to a function instead of a variable.
I should have mentioned this in my original post, but part of my confusion is also coming from a book I have (Classical Dynamics by Thornton and Marion), which says:
\frac{\partial \mathcal{S}}{\partial \alpha} d \alpha \equiv \delta \mathcal{S}
where
\mathcal{S}(\alpha) = \int_{t_1}^{t_2} \! L(\mathbf{q}(\alpha, t), \dot{\mathbf{q}}(\alpha, t); t) \, dt
and
\mathbf{q}(\alpha, t) = \mathbf{q}(0, t) + \alpha \eta(t)
and
\eta(t) is a function such that \eta(t_1) = \eta(t_2) = 0
I'm having trouble seeing how these two definitions of the \delta operator are equivalent. Thornton and Marion's \eta(t) seems to be the same as Wikipedia's \boldsymbol\varepsilon(t).
Does the following look correct?
Let f be a function f(x(t)).
Approaching from the Wikipedia side:
Let \varepsilon(t) be a small perturbation from x(t).
\delta f = f(x(t) + \varepsilon(t)) = \varepsilon(t) \frac{\partial f}{\partial x(t)}
Approaching from the Thornton and Marion side:
x(t) = x(0, t); x(\alpha, t) = x(0, t) + \alpha \eta(t)
\delta f = \frac{\partial f}{\partial \alpha} d \alpha = \frac{\partial f}{\partial x(t)} \frac{\partial x(t)}{\partial \alpha} d \alpha = \eta(t) \frac{\partial f}{\partial x(t)} d \alpha
I'm not sure what to do with that d \alpha in the Thornton and Marion approach. Should that be there?
