General Chain Rule Product Rule:
For a two $\mathcal{C}^{\infty}(\mathbb{R})$ functions, $u(x)$, $v(x)$ we have
$$\frac{d^k}{dx^{k}}[u(x)v(x)]=\sum_{j=0}^k \binom{k}{j}\frac{d^j}{dx^j}[u(x)]\frac{d^{k-j}}{dx^{k-j}}[v(x)]$$
Now, I am interested in an analogue for the chain rule, consider $g(t)=f(u(t))$
I began working with low order derivatives to see if I could see a pattern emerge that I could then prove holds rigorously but have had no such luck. The result seems to be a classical one but will aid in my research.
I can see the "end terms" of a kth order derivative expansion
$$\frac{d^k}{dt^k} f(u(t))=\frac{d^k}{du^k}[f(u)]\left[\frac{du}{dt}\right]^k+\cdots+\frac{d}{du}[f(u)]\frac{d^k}{dt^k}[u(t)]$$
 A: From https://en.wikipedia.org/wiki/Faà_di_Bruno%27s_formula:

Faà di Bruno's formula is an identity in mathematics generalizing the chain rule to higher derivatives, named after Francesco Faà di Bruno, though he was not the first to state or prove the formula. In 1800, more than 50 years before Faà di Bruno, the French mathematician Louis François Antoine Arbogast stated the formula in a calculus textbook, considered the first published reference on the subject.
Perhaps the most well-known form of Faà di Bruno's formula says that
$${d^n \over dx^n} f(g(x))=\sum \frac{n!}{m_1!\,1!^{m_1}\,m_2!\,2!^{m_2}\,\cdots\,m_n!\,n!^{m_n}}\cdot f^{(m_1+\cdots+m_n)}(g(x))\cdot \prod_{j=1}^n\left(g^{(j)}(x)\right)^{m_j},$$
where the sum is over all n-tuples of nonnegative integers $m_1,\ldots,m_n$ satisfying the constraint
$$1\cdot m_1+2\cdot m_2+3\cdot m_3+\cdots+n\cdot m_n=n.\,$$
Sometimes, to give it a memorable pattern, it is written in a way in which the coefficients that have the combinatorial interpretation discussed below are less explicit:
$${d^n \over dx^n} f(g(x))
=\sum \frac{n!}{m_1!\,m_2!\,\cdots\,m_n!}\cdot
f^{(m_1+\cdots+m_n)}(g(x))\cdot
\prod_{j=1}^n\left(\frac{g^{(j)}(x)}{j!}\right)^{m_j}.$$
Combining the terms with the same value of $m_1 + m_2 + \cdots + m_n = k$ and noticing that $m_j$ has to be zero for $j > n − k + 1$ leads to a somewhat simpler formula expressed in terms of Bell polynomials $B_{n,k}(x_1,\ldots,x_{n-k+1})$:
$${d^n \over dx^n} f(g(x)) = \sum_{k=1}^n f^{(k)}(g(x))\cdot B_{n,k}\left(g'(x),g''(x),\dots,g^{(n-k+1)}(x)\right).$$

