Is there a compact expression for multiple chain rule derivatives? Let $h(x) = f(g(x))$. Is there an expression for $h^{(n)}(x)$, which denotes the $n^{\text{th}}$ derivative of $h$ with respect to $x$, in terms of derivatives of $f$ and $g$?
For example, the first few derivatives are easily checked to be:
$h'(x) = f' \cdot g'$
$h''(x) = f'' \cdot(g')^2 + f' \cdot g''$
$h'''(x) = f''' \cdot(g')^3 + 3f'' \cdot (g')^2\cdot g''+f'\cdot g'''$
$h^{(4)}(x)=f^{(4)} \cdot(g')^4 + 6f'''\cdot (g')^2\cdot g'' + 3f''\cdot(g'')^2+4g''\cdot f'\cdot f'' +g'\cdot f^{(4)}, \\$
where $f^{(n)}$ is understood to mean $f^{(n)}(g(x))$.

One might expect a nice pattern to emerge, but I'm not seeing anything obvious. Compare this to the Pascal's triangle pattern that emerges for $h(x)=f(x)\cdot g(x)$ upon taking $n$ derivatives of $h$(x), allowing us to write very simply, $h^{(n)}(x)=\sum_{i=0}^n {{n}\choose{i}}f^{(n-i)}(x)\cdot g^{(i)}(x)$.
 A: A formula to calculate the $n$-th derivative of composite functions is stated as identity (3.56) in H.W. Gould's Tables of Combinatorial Identities, Vol. I and called:

Hoppe Form of Generalized Chain Rule
Let $D_g$ represent differentiation with respect to $g$ and $g=g(x)$. Hence $D^n_x f(g)$ is the $n$-th derivative of $f$ with respect to $x$.  The following is valid for $n\geq 1$:
\begin{align*}
D_x^n f(g)=\sum_{k=1}^nD_g^kf(g)\frac{(-1)^k}{k!}\sum_{j=1}^k(-1)^j\binom{k}{j}g^{k-j}D_x^ng^j\tag{1}
\end{align*}

Let's look at a small  example $n=2$:
Differentiating twice we obtain
\begin{align*}
\left(f(g(x))\right)^{\prime\prime}&=\left(f^{\prime}(g(x))g^{\prime}(x)\right)^{\prime}\\
&=f^{\prime\prime}\left(g(x)\right)\left(g^{\prime}(x)\right)^2+f^{\prime}(g(x))g^{\prime\prime}(x)\tag{2}
\end{align*}
Using Hoppe's formula (1) we obtain
\begin{align*}
\color{blue}{D_x^2f(g)}&=D_gf(g)\frac{(-1)}{1!}\left((-1)\binom{1}{1}g^0D_x^2g\right)\\
&\qquad+D_g^2f(g)\frac{(+1)}{2!}\left((-1)\binom{2}{1}g^1D_x^2g+\binom{2}{2}g^0D_x^2g^2\right)\\
&=f^{\prime}\left(g(x)\right)\left(g^0(x)g^{\prime\prime}(x)\right)
+f^{\prime\prime}(g)\frac{1}{2}\left(-2g(x)g^{\prime\prime}(x)+g^0(x)D_x\left(2g(x)g^{\prime}(x)\right)\right)\\
&=f^{\prime}\left(g(x)\right)\left(g^0(x)g^{\prime\prime}(x)\right)
+f^{\prime\prime}(g)\frac{1}{2}\left(-2g(x)g^{\prime\prime}(x)+2\left({g^{\prime}}(x)\right)^2+2g(x)g^{\prime\prime}(x)\right)\\
&\,\,\color{blue}{=f^{\prime}\left(g(x)\right)g^{\prime\prime}(x)+f^{\prime\prime}\left(g(x)\right)\left(g^{\prime}(x)\right)^2}
\end{align*}
in accordance with (2).
