Why is Cartan Formula just an avatar of Leibniz rule? In this video, Arnold says that the Cartan formula $$ \mathscr L_{\mathrm X} = d i_{\mathrm X} + i_{\mathrm X} d$$
is just an avatar of $(fg)' = fg' + f'g$.
Why ?
 A: Wow. This video reminds me of how sad I am not to have heard Arnol'd speak while he was alive.
The following is an attempt to work out what he might have meant by the observation.
Suppose that $V$ is a smooth vector field on a manifold $M$.
Let $C$ be a smooth $k$-chain in $M$ and let $\omega$ be a smooth $k$-form on $M$. Denote the flow of $V$ by $\phi_t$.
Let $\Phi$ be the embedding of $I \times C \to M$ by $(t, x) \mapsto \phi_t(x)$, where $I = [0, T]$ is an interval and $t$ is the coordinate on it.
For any $(k+1)$-form $\Omega$ on $M$, we have $\Phi^*\Omega = \phi_t^*\Omega + dt \wedge i_V \Omega$ on $I \times C$.
Now, the boundary of $I \times C$ is
$$-(\{0\} \times C) \cup I \times \partial C \cup \{T\} \times C.$$
If the signs are done correctly, this is the formula Arnol'd also calls the Leibniz formula for a Cartesian product of chains, that
$\partial (A\times B) =\partial A \times B \cup A \times \partial B$.
Now, by Stokes' theorem
$$
\int_{I \times C} \Phi^*d\omega = -\int_{0 \times C} \Phi^*\omega + \int_{I \times \partial C} \Phi^*\omega + \int_{T \times C} \Phi^*\omega.
$$
If we plug in what we computed for $\Phi^*\Omega$
(on the LHS, $\Omega = d\omega$ and on the RHS, $\Omega = \omega$),
we get
$$
 \int_{I \times C} \phi_t^*d\omega + dt \wedge i_V d\omega
= 
- \int_{0 \times C} \omega 
+ \int_{I \times \partial C} dt \wedge i_V \omega
+ \int_{T \times C} \phi_T^* \omega.
$$
Let me first explain why, from this step, you get Cartan's magic formula.
Differentiate both sides with respect to $T$. This becomes
$$
\int_{C} d\omega + i_V d\omega = -0 + 0 + \int_C L_V \omega.
$$
This is true for an arbitrary smooth chain $C$ so the equality holds at the level of $k$-forms pointwise.
Now, the reason this is the same proof is that we have drawn the same picture. Indeed, in both cases, the key step is to use the picture he drew on the board of $I \times C$ and determine its boundary, and differentiate this with respect to the $t$ parameter.

EDIT: here is a proof that
$\Phi^*\Omega = \phi_t^*\Omega + dt \wedge i_V \Omega$ on $I \times C$.
Note first that by definition $\Phi^*\Omega|_{t,x} = \Omega( d\Phi(t,x) \cdot , d\Phi(t,x) \cdot, \dots, d\Phi(t,x) \cdot)$. Now, $d\Phi(t,x) = d\phi_t(x) + V(\phi_t(x)) \otimes dt$. In other words, $d\Phi(t,x) \, v = d\phi_t(x)\,v$ for each $v \in TC$ and $d\Phi(t,x)\,\partial_t = V(\phi_t(x))$.
Unenlightening proof 1: combine this expression for $d\Phi$ with $\Omega$ and you get the formula I claim.
Proof 2: use local coordinates. Choose a basis for $TC$ at the point $x$.
We have that $\Phi^*\Omega$ is a form on $C \times I$, so we can write it as $\mu + dt \wedge \nu$ for $t$-dependent forms $\mu$ and $\nu$ on $I \times C$.
To compute $\mu$, we evaluate $\Phi^*\Omega$ on the basis of $TC$, and get $\mu = \phi_t^*\Omega$.
To compute $\nu$, we evaluate $\Phi^*\Omega$ on $\partial_t$ followed by a collection of vectors from our basis of $T_xC$. This gives us the $i_V \Omega$ term.
It would be great if someone could teach me a nice proof of this fact, actually.

OK, as I fill in more details, the link between the two proofs becomes more and more tenuous. I suspect that Arnol'd is saying something deeper than the similarity I am noticing. I think I tend to fall back into the more "algebraic" way of thinking of the exterior derivative, and he has a more geometric picture in mind.
