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

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Go to the Online Encyclopedia of Integer Sequences (OEIS) on the Net and search under Lagrange inversion and also series reversion and you will find many examples. –  Tom Copeland Jun 1 '12 at 7:48

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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.

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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

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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.

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This is sometime called the Lagrange inversion formula.

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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.

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