Higher derivatives of the map $I:T \mapsto T^{-1}$, where $T \in \mathcal B(X)$.

For small enough $t$ we have the power series expansion

$$ f(T + tS) = (T + tS)^{-1} = (T(1 - (-T^{-1}S)))^{-1} = T^{-1} \sum_{k=0}^{\infty} (-1)^k (T^{-1}S)^k t^k. $$

Using it we see that,

$$ (D^k f)|_{T}(S, \dots, S) = \left( \frac{d}{dt} \right)^k f(T + tS)|_{t=0} = k! (-1)^k (T^{-1} S)^k. $$

Now we know that $D^k f|_{T}(S_1,\dots,S_k)$ is symmetric and uniquely determined from $D^k f|_{T}(S,\dots,S)$ by the polarization identity so we can just guess that $$ D^k f|_{T}(S_1,\dots,S_k) = (-1)^k \sum_{\sigma \in S_k} T^{-1} S_{\sigma(1)} \cdots T^{-1} S_{\sigma(k)} $$

and since this is symmetric in $S_1,\dots,S_k$ and coincides with our expression when $S_1 = \dots = S_k = S$, our guess must hold.