How to find the derivative of the adjugate operator? How can I find the general derivative of the function $\mathbf{A} \to \mathrm{adj}(\mathbf{A})$ ?
Where $\mathbf{A}$ is invertible I got the following:
$\mathrm{D}f_{\mathrm{adj}(\mathbf{A})(H)} = \mathrm{tr}(\mathrm{adj}(\mathbf{A}) \cdot H)\cdot \mathrm{adj}(\mathbf{A})^{-1}   -   (\mathrm{adj}(\mathbf{A})) \cdot  H \cdot\mathrm{adj}(\mathbf{A})^{-1}$
Is it right? (And if it is, it's still not enough for every $\mathbf{A}$..)

If $A$ is invertible, then $f(A)=adj(A)=\det(A)A^{-1}$. Then $Df_A:H\rightarrow tr(adj(A)H)A^{-1}-\det(A)A^{-1}HA^{-1}=tr(adj(A)H)A^{-1}-adj(A)HA^{-1}$.
EDIT. There was a mistake in this part.
If $A$ is not invertible, it is more complicated and I give only an idea. $H$ is given and $K=Df_A(H)$ is unknown. One has $A.adj(A)=adj(A).A=\det(A).I=0$ but, when $\det(A)=0$, these relations do not characterize $adj(A)$; moreover , the derivative of RHS is not $0$.
We obtain $(*)$ $H.adj(A)+AK=KA+adj(A).H=tr(adj(A)H).I$ but these relations do not characterize $K$. 
Note that, when $rank(A)\leq n-2$, $adj(A)=0$ and the previous relations reduce to $AK=KA=0$.
When $rank(A)=n-1$, $rank(adj(A))=1$ and the solutions $K$ of $(*)$ are in an affine space of dimension $1$, that is, $K=u+tv$ where $u,v$ are known vectors (dependent on $H$) and $t$ is an unknown scalar.
