Faà di Bruno's formula for inverse functions? It is easy to get a expression for the nth-derivative of an inverse function ; starting from $(f^{-1})'=\frac{1}{f'\circ f^{-1}}$, one gets things like $(f^{-1})^{(n)}=\frac{\sum a_k\prod (f^{(n_j)}\circ f^{-1})^j}{(f'\circ f^{-1})^{2n-1}}$, with reasonably easy constraints on the $n_j$. But what are the values of the $a_k$? I believe I read somewhere this was an application of umbral calculus, but I don't see how, and inverting Faà di Bruno's formula on the identity $f\circ f^{-1}=id$ don't seem to get anywhere.
 A: To precise my question, I was asking for the exact values of the $a_k$. Thanks to Tom Copeland, I could find the sequence A176740 of OEIS, giving a complete answer (with useful links) to this problem.
A: Since this seems to one of those perennial questions for which people propose the Faà di Bruno (Bell–Touchard–Scherk-refined Stirling) partition polynomials as a route to a solution and don't attempt to look at the OEIS for answers, I thought I would amplify on my 2012 comment, via which the OP found some answers to his question, by listing the pertinent OEIS entries and some related MO-Qs and noting that the refined Lah partition polynomials, a close cousin of the Faà-di-Bruno partition polynomials, provide the most elegant rep of the inversion polynomials using umbral notation.
OEIS entries on families of Lagrange compositional inversion (reversion) partition polynomials (IPPs):
Let $g(z) = 1/f'(z)$. The first several polynomials of different families of IIPs expressed in various indeterminates can be generated with a symbolic math app, like Mathcad, by expanding the Lie action $(g(s)\partial_z)^n g(z)$ with different reps for $g(z)$ as done for the following OEIS entries;

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*A145271, $g(z) = g_0 + g_1 z + g_2 \frac{z^2}{2!}+g_3 \frac{z^3}{3!}+\dotsb $, generic $\mathit{Sl}_n$ version, refined Eulerian numbers

*A134685 and A176740, $g(z) = 1/\partial_z ( a_1 z + a_2 \frac{z^2}{2!}+a_3 \frac{z^3}{3!}+\dotsb)$, e.g.f. rep

*A133437,  $g(z) = 1/\partial_z ( b_1 z + b_2 z^2+b_3 z^3+\dotsb)$, o.g.f. rep

*A133932, $g(z) = 1/\partial_z (c_1 z + c_2 \frac{z^2}{2}+a_3 \frac{z^3}{3}+\dotsb)$, log rep,

*A134264, $g(x)=  1/\partial_z (z/(h_0 + h_1 z + h_2 z^2+h_3 z^3+\dotsb))$, reciprocal o.g.f. rep

*A248120, $g(x)=  1/\partial_z (z/(q_0 + q_1 z + q_2 \frac{z^2}{2!}+q_3 \frac{z^3}{3!}+\dotsb))$ reciprocal e.g.f. rep

*A248927, simply scaled version of 6).

Direct multinomial-type formulas are presented in the OEIS entries and links therein for the numerical coefficients of the partition monomials / summands of the IPPs of entries 2–7. The oldest family is, most likely, 3), developed by Newton, and later recognized first, most likely, by Loday as the refined Euler characteristic partition polynomials of the celebrated associahedra convex polytopes. An umbral representation of the associahedra family of IPPs is given on p. 11 of my notes "Lagrange à la Lah--Lagrange Landscapes Part I: Composition and Inversion
via Partition Polynomials" at my blog post "Lagrange à la Lah", using the refined Lah compositional partition polynomials rather than its close cousin the Faà-di-Bruno compositional partition polynomials. Diverse combinatorial constructs are linked to the IPPs—I've already mentioned the associahedra, and another famous one is the family of noncrossing partitions, both have multiple combinatorial interpretations.
Related MO-Qs with links to others:
I) "A Leibniz-like formula for $(f(x)\frac{d}{dx})^nf(x)$?"
II) "Formula for n-th iteration of dx/dt=B(x)".
A: You should be able to get a formula, first by reducing to the case where f(0)=0
and the evaluation of the derivatives (for both f and its inverse) is at 0.
Then, work formally by replacing f by its Taylor-MacLaurin series at 0. The problem
then becomes that of the reversion of power series. It has been done in many places and
typically involves summing over trees.
A: This is sometime called the Lagrange inversion formula.
A: The question can be reduced to a problem involving power series.
In this direction, see Proposition 1.1 (and the preceding discussion) of this paper:

*

*Samuel G. G. Johnston, Joscha Prochno, Faà di Bruno's formula and inversion of power series, Advances in Mathematics
395 (2022) 108080, https://doi.org/10.1016/j.aim.2021.108080, https://arxiv.org/abs/1911.07458
It states that if $f(s) = s + \sum_{j=2}^\infty a_j s^j$ is a formal power series, then the formal inverse power series $f^{-1}$ has coefficients which can be expressed in terms of weighted rooted trees.
A: See Warren P. Johnson, Combinatorics of Higher Derivatives of Inverses,
American Mathematical Monthly,
Vol. 109, No. 3 (Mar., 2002), pp. 273-277,
http://www.jstor.org/stable/2695356
A: Riordan's Combinatorial identities has a chapter on partition polynomials that may be helpful. It specifically covers the question you are asking, but is in umbral calculus. 
A: I. G. Macdonald gives a very explicit formula for the coefficients in Example 24, p35 of the 2nd Edition of Symmetric Functions and Hall Polynomials. The example begins `Another involution on the ring $\Lambda$ may be defined as follows...'
A: My answer contains partial exponential Bell polynomials.
Because the problem concerns higher derivatives of compositions of functions, it can be answered by using partial exponential Bell polynomials.
Set $f^{-1}=\phi$.
Let $B$ denote partial exponential Bell polynomials.
Applying Faà di Bruno's formula (higher chain rule of differentiation) to the inverse rule of differentiation $\phi'=\frac{1}{f'\circ \phi}$, we get
$$\phi^{(n)}=\sum_{k=0}^{n-1}(-1)^kk!(f'\circ\phi)^{-(k+1)}B_{n-1,k}(f'\circ\phi).$$
That means, we still need an expression for $B_{n-1,k}(f'\circ\phi)$.
$\ $
The usual way is to use Lagrange inversion theorem instead. Lagrange inversion theorem gives the Taylor series expansion of the inverse function of an analytic function.
The inverse of an analytic function $f$ in a neighbourhood of a point $a$ can be written as a power series iff $f'(a)\neq 0$.
For $f(x)=\sum_{n=0}^\infty\frac{f_n}{n!}x^n$, $\phi(x)=\sum_{n=0}^\infty\frac{\phi_n}{n!}x^n$ and $f_0=0$, Lagrange inversion theorem gives
$$\phi_n=(n-1)!\left[x^{n-1}\right]\left(\frac{x}{f(x)}\right)^n.$$
This is sequence A176740 in the Online Encyclopedia of Integer Sequences (OEIS). See the other answers.
Clearly you can derive this also with umbral algebra.
$\ $
Let's look at the representation with Bell polynomials.
Applying Faà di Bruno's formula to $\left(\frac{x}{f(x)}\right)^n$ gives for $n\ge 2$
$$\phi_n=(n-1)!\frac{1}{f_1^n}\sum_{k=1}^{n-1}(-1)^k(n)_kB_{n-1,k}\left(\frac{f_2}{2f_1},\frac{f_3}{3f_1},...,\frac{f_{n-k+1}}{(n-k+1)f_1}\right)$$
with $(n)_k$ the rising factorial (Pochhammer symbol).
This is already written in the Wikipedia article for Lagrange inversion theorem.
$$B_{n,k}(f_1,f_2,...,f_n)=\sum_{1k_1+2k_2+...+(n-k+1)k_{n-k+1}=n\atop k_1+k_2+...+k_{n-k+1}=k}\frac{n!}{1!^{k_1}k_1!2!^{k_2}k_2!\cdot ...\cdot (n-k+1)!^{k_{n-k+1}}k_{n-k+1}!}f_1^{k_1}f_2^{k_2}\cdot ...\cdot f_{n-k+1}^{k_{n-k+1}}$$
$$B_{n-1,k}\left(\frac{f_2}{2f_1},\frac{f_3}{3f_1},...,\frac{f_{n-k}}{(n-k)f_1}\right)$$
$$=\sum_{1k_1+2k_2+...+(n-k)k_{n-k}=n-1\atop k_1+k_2+...+k_{n-k}=k}\frac{(n-1)!}{1!^{k_1}k_1!2!^{k_2}k_{2}!\cdot ...\cdot (n-k)!^{k_{n-k}}k_{n-k}!}\left(\frac{f_2}{2f_1}\right)^{k_1}\left(\frac{f_3}{3f_1}\right)^{k_2}\cdot ...\cdot \left(\frac{f_{n-k}}{(n-k)f_1}\right)^{k_{n-k}}$$
$$=\frac{1}{f_1^k}\sum_{1k_1+2k_2+...+(n-k)k_{n-k}=n-1\atop k_1+k_2+...+k_{n-k}=k}\frac{(n-1)!}{1!^{k_1}k_1!2!^{k_2}k_2!\cdot ...\cdot (n-k)!^{k_{n-k}}k_{n-k}!}\frac{1}{2^{k_1}3^{k_2}(n-k)^{k_{n-k}}}f_2^{k_1}f_3^{k_2}\cdot ...\cdot f_{n-k}^{k_{n-k}}$$
Now we have an explicit formula for $\phi_n$.
We could also explore $B_{n-1,k}\left(\frac{f_2}{2f_1},\frac{f_3}{3f_1},...,\frac{f_{n-k}}{(n-k)f_1}\right)$ to try to get some recurrence formulas.
A: I believe that it's Comtet's Advanced Combinatorics that has a chapter on partition polynomials. A section explains how to find the inverse of a function using $f^{-1}(f(x))=1$. So the solution involves composition and therefore is associated with set partitions being the combinatorial. My work has a stronger result that the partition polynomials and Faà di Bruno's formula allows for iterated functions to be derived.
