I am currently trying to understand the BV-formalism, which makes heavy use of the functional derivative.

Let us consider the functional derivative, as defined in for example its Wikipedia article. Let $F$ be a functional, i.e. a map from, say, $C^\infty(\mathbb{R})$ to $\mathbb{R}$, and suppose it may be written as $F[\phi] = \int f\big(x,\phi(x),\phi'(x),\dots,\phi^{(n)}(x)\big)\\,dx$ for some function $f$ which depends on the derivatives of $\phi$ up to order $n$. Then the functional derivative of $F$ is $\displaystyle \frac{\delta F}{\delta \phi} = \sum_{i=1}^n(-1)^i\frac{d^i}{dx^i}\frac{\partial f}{\partial \phi^{(i)}}$.

Now, my background is that of differential equations and differential geometry, i.e. jet spaces and variational calculus and the like. In that area, the latter operator, $\sum_{i}(-1)^i\frac{d^i}{dx^i}\frac{\partial}{\partial \phi^{(i)}}$, is well known; it is called the variational derivative. Summarizing, then, we seem to have that the functional derivative of a functional is the variational derivative of (one of its) densities.

Since the variational derivative involves lots of derivatives, it certainly does not satisfy the Leibniz rule, i.e. it is not a derivation. In various places, however, I've come across the statement that the functional derivative does satisfy the Leibniz rule. (That already seems unexpected to me: how can an operator which is so intimately connected to a decidedly non-derivation be a derivation?) There are various ways to prove it, but I would like to understand this fact in terms of the variational derivative, if possible. So: how can the Leibniz rule of the functional derivative related to variational derivative; can the former be expressed somehow in terms of the latter?

I am currently trying to understand the BV-formalism, which makes heavy use of the functional derivative.

Let us consider the functional derivative, as defined in for example its Wikipedia article. Let $F$ be a functional, i.e., a map from, say, $C^\infty(\mathbb{R})$ to $\mathbb{R}$, and suppose it may be written as $F[\phi] = \int f\big(x,\phi(x),\phi'(x),\dots,\phi^{(n)}(x)\big)dx$ for some function $f$ which depends on the derivatives of $\phi$ up to order $n$. Then the functional derivative of $F$ is $\displaystyle \frac{\delta F}{\delta \phi} = \sum_{i=1}^n(-1)^i\frac{d^i}{dx^i}\frac{\partial f}{\partial \phi^{(i)}}$.

Now, my background is that of differential equations and differential geometry, i.e., jet spaces and variational calculus and the like. In that area, the latter operator $\sum_{i}(-1)^i\frac{d^i}{dx^i}\frac{\partial}{\partial \phi^{(i)}}$ is well known; it is called the variational derivative. Summarizing, then, we seem to have that the functional derivative of a functional is the variational derivative of (one of its) densities.

Since the variational derivative involves lots of derivatives, it certainly does not satisfy the Leibniz rule, i.e., it is not a derivation. In various places, however, I've come across the statement that the functional derivative does satisfy the Leibniz rule. (That already seems unexpected to me: how can an operator which is so intimately connected to a decidedly non-derivation be a derivation?) There are various ways to prove it, but I would like to understand this fact in terms of the variational derivative, if possible. So: how can the Leibniz rule of the functional derivative related to variational derivative; can the former be expressed somehow in terms of the latter?