It is easy to get a expression for the nthderivative of an inverse fuction ; starting from $(f^{1})'=\frac{1}{f'\circ f^{1}}$, one gets things like $(f^{1})^{(n)}=\frac{\sum a_k\prod (f^{(n_j)}\circ f^{1})^j}{(f'\circ f^{1})^{2n1}}$, 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 dont see how, and inverting Faa di Bruno's formula on the identity $f\circ f^{1}=id$ dont seem to get anywhere.

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. 


See Warren P. Johnson, Combinatorics of Higher Derivatives of Inverses, American Mathematical Monthly, Vol. 109, No. 3 (Mar., 2002), pp. 273277, http://www.jstor.org/stable/2695356 


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 TaylorMacLaurin 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. 


This is sometime called the Lagrange inversion formula. 


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. 


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...' 

