 Solution

For small $\Vert \Delta \Vert$, 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 $\Vert A^{-1}\Delta \Vert <1$. Applying the resolvent identity twice givesThe last term is $O(\Vert \Delta {\Vert }^2)$ by that local bound. Hence the inverse map is differentiable, withThis verifies the differentiability and its remainder, rather than only formally differentiating the identity $A{A}^{-1}=I$.