Faà di Bruno's formula for multiple arguments but still with respect to one variable Faà di Bruno's formula given (from Wikipedia) by
.
The following are questions about generalizations that are similar, but not the same:

*

*Multivariate Faà di Bruno's formula

*Extension Faà di Bruno formula

*Faa di Bruno's formula generalization
What I would like an expression for is $$\frac{d^n}{dx^n} f(g_1(x), \ldots, g_k(x))$$ which involves taking the $n$th order derivative with respect to only $x$ but there are multiple arguments of $f$.
 A: Consider $f(a(x),b(x))$. Then
$$f_x=f_aa_x+f_bb_x$$
$$f_{xx}=f_{aa}a_x^2+2f_{ab}a_xb_x+f_{bb}b_x^2+f_aa_{xx}+f_bb_{xx}$$
$$f_{xxx}=\color{blue}{f_{aaa}a_x^3 + 3f_{aab}a_x^2b_x + 3f_{abb}a_xb_x^2 + f_{bbb}b_x^3}
+\color{black}{3(f_{aa}a_{xx}a_x + f_{ab}a_{xx}b_x + f_{ab}b_{xx}a_x + f_{bb}b_{xx}b_x)}
+\color{blue}{f_aa_{xxx} + f_bb_{xxx}}$$
You can see where this is going: if $f$ has $r$ arguments, for each $f_{a^k}a_{x^{n_1}}a_{x^{n_2}}\cdots a_{x^{n_k}}$ corresponding to a partition in the classic formula ($n_1+\cdots+n_k$ is always the derivative order desired) we replace it with a sum of $r^k$ terms formed by all possible ways of "colouring" the $k$ pairs of $f$-differentiating variables and $a_{x^{n_i}}$ factors with the $r$ available arguments. As a further example, consider the term corresponding to the $2+1+1$ partition in $f^{(4)}(g(x))$:
$$6(f'''(g(x))g''(x)g'(x)^2)$$
Except for the $6$, this is replaced by a sum of $8$ terms in $f_{xxxx}(a(x),b(x))$:
$$\begin{align}&f_{aaa}a_{xx}a_xa_x\\
+&f_{aab}a_{xx}a_xb_x\\
+&f_{aba}a_{xx}b_xa_x\\
+&f_{abb}a_{xx}b_xb_x\\
+&f_{baa}b_{xx}a_xa_x\\
+&f_{bab}b_{xx}a_xb_x\\
+&f_{bba}b_{xx}b_xa_x\\
+&f_{bbb}b_{xx}b_xb_x\end{align}$$
