= Solution
{c}
For small $\|\Delta\|$, the <matrix> $A+\Delta=A(I+A^{-1}\Delta)$ remains invertible and its inverse is locally bounded, for example by the convergent <Neumann series> when $\|A^{-1}\Delta\|<1$. Applying the <resolvent identity> twice gives
$$
(A+\Delta)^{-1}
=A^{-1}-A^{-1}\Delta A^{-1}
+A^{-1}\Delta A^{-1}\Delta(A+\Delta)^{-1}.
$$
The last term is $O(\|\Delta\|^2)$ by that local bound. Hence the inverse map is differentiable, with
$$
\boxed{D\mathcal I_A(\Delta)=-A^{-1}\Delta A^{-1}.}
$$
This verifies the differentiability and its remainder, rather than only formally differentiating the identity $AA^{-1}=I$.
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