Imagine you have a smooth functional $ F : M \to \mathbb{R} $ where $ M $ is the space of smooth functions $ s : [a,b] \to \mathbb{R} $ satisfying certain boundary conditions. Assume also that in the space $M$ you have a smooth curve $ \alpha : \mathbb{R} \to M $.
How would you compute the derivative $$ \tfrac{d}{dt}(F \circ \alpha) (t) $$ for any $t \in \mathbb{R}$?
In particular, I am interested in the case when the curve $\alpha$ gives you a critical point of $F$ for every $t$. Intuition tells me that the derivative $\tfrac{d}{dt}(F \circ \alpha) (t)$ should be zero, but I am not able to find a formal way of writing it.
I thought of using something like a "chain rule" for functionals and then using the fact that $\alpha (t)$ is a critical point of $F$, but I know this is delicate so I am not so sure. Any idea?
Maybe consider an extension $F:C^\infty([a, b]) \rightarrow \mathbb{R}$. Then endow $F$ with a norm such that $F$ is Fréchet-differentiable (definition works basically like $\mathbb{R}^n$, you can look it up everywhere on the internet). We do not know whether such norm exists but we wanted to assume regularity so there we go. The derivative in any $s \in C^\infty([a, b])$ is then a linear, continuous operator $DF_s:C^\infty([a, b]) \rightarrow \mathbb{R}$.
You can prove a chain rule for Fréchet-derivatives exactly the way you do for $\mathbb{R}^n$. So we get for any $t$: $$ \frac{\mathrm{d}}{\mathrm{d}t} F(\alpha(t)) = DF_\alpha (\alpha'(t)) $$ So since $\alpha$ is a critical point, the linear operator $C^\infty([a, b]) \ni v \mapsto DF_\alpha (v)$ is zero. So yes, your desired derivative is zero. If you need further detail, let me know.