classical mechanics - Does the variation of the Lagrangian satisfy the product rule and chain rule of the derivative?

I have seen wikipedia use the product rule and maybe the chain rule for the variation of the Langragin as follows:

$$\begin{align} \dfrac{\delta [f(g(x,\dot{x}))h(x,\dot{x})] } {\delta x} = \left( \dfrac{\delta [f(g)] } {\delta g} \dfrac{\delta [g(x,\dot{x})] } {\delta x} \right) h(x,\dot{x}) + f(g(x,\dot{x})) \dfrac{\delta [h(x,\dot{x})] } {\delta x} \end{align}$$ where the variation of the Lagrangian is defined $$\begin{align} \dfrac{\delta \mathcal{L} } {\delta x} = \dfrac{\partial \mathcal{L} } {\partial x} - \dfrac{d}{d \tau} \dfrac{\partial \mathcal{L} } {\partial \dot{x}} \end{align}$$ and $\mathcal{L}=f(g(x,\dot{x}))h(x,\dot{x})$.

Does the variation of the Lagrangian satisfy the product rule and chain rule of the derivative?

Answer

1. 
   

   OP considers the 'same-time' functional derivative (FD)

   $$\tag{1} \frac{\delta f(t)}{\delta x(t)}~:=~\frac{\partial f(t)}{\partial x(t)} - \frac{d}{dt} \frac{\partial f(t)}{\partial \dot{x}(t)} +\ldots.$$ Here $f(t)$ is shorthand for the function $f(x(t), \dot{x}(t), \ldots;t)$. Although the 'same-time' FD (1) can be notationally useful, it has various fallacies, cf. my Phys.SE answer here.
   
   


2. 
   

   The Leibniz rule

   $$\tag{2} \frac{\delta (f(t)g(t))}{\delta x(t)} ~=~\frac{\delta f(t)}{\delta x(t)} g(t) +f(t)\frac{\delta g(t)}{\delta x(t)}\qquad(\leftarrow \text{Wrong!})$$ for the 'same-time' FD (1) does not hold. Counterexample: Take $f(t)=g(t)=\dot{x}(t)$.
   
   


3. 
   

   The chain rule

   $$\tag{3} \frac{\delta f(t)}{\delta x(t)} ~=~\frac{\delta f(t)}{\delta y(t)}\frac{\delta y(t)}{\delta x(t)}\qquad\qquad(\leftarrow \text{Wrong!})$$ for the 'same-time' FD (1) does not hold. Counterexample: Take $f(t)=y(t)^2$ and $y(t)=\dot{x}(t)$.
   
   
   


4. 
   

   However, the usual FD $\frac{\delta F}{\delta x(t)}$ (where $F[x]$ is a functional) does satisfy a Leibniz rule

   $$\tag{4} \frac{\delta (FG)}{\delta x(t)} ~=~\frac{\delta F}{\delta x(t)} G +F\frac{\delta G}{\delta x(t)},$$ and a chain rule $$\tag{5} \frac{\delta F}{\delta x(t)}~=~ \int dt^{\prime} ~\frac{\delta F}{\delta y(t^{\prime})}\frac{\delta y(t^{\prime})}{\delta x(t)}.$$