calculus and analysis - On the n-th derivative of the inverse function

Given a function $f(x)$, its inverse $g(x)$ is defined as $g(f(x)) \equiv x$.

In light of this, the n-th derivative of $g(x)$ can be recursively calculated as follows:

list = {};

For[n = 1, n <= 6, n++,

    eqn = D[g[f[x]], {x, n}] == D[x, {x, n}];

    var = Derivative[n][g][f[x]];

    If[n == 1,
       list = Join[list, Solve[eqn, var][[1]]],
       list = Join[list, (Solve[eqn, var] /. list)[[1]]]
      ]

   ]

list // Expand // TableForm

In MMA you can get this very simply by writing something like this:

Derivative[n][InverseFunction[f]][f[x]]

The question that arises is the following: is the latter formulation using a recursive process like the one shown above or is there a non-recursive formulation?

Answer

D[InverseFunction[f][x], {x, 6}]

Towards the question whether this is computed recursively: I guess so from analyzing the Trace produced by executing the code:

Table[
 Length@Trace[
   D[InverseFunction[f][x], {x, k}]
   ],
 {k, 1, 12}]

{6, 7, 7, 10, 13, 18, 25, 36, 51, 73, 103, 145}