It is well known how to take the derivative of the determinant: let $A(s)$ be a family of square matrices smoothly parametrised by the variable $s$ (in other words, $A:\mathbb{R}\to \mathbb{R}^{N^2}$ is a smooth curve). Then the derivative of the determinant can be given by
\frac{d}{ds} \det A(s) = \operatorname{Tr} \left(\tilde{A}(s) A'(s)\right)
where A'(s) = \frac{d}{ds}A(s) and $\tilde{A}(s)$ is the adjugate matrix of $A(s)$. The above formula also generalises to the case that $A:\mathbb{R}^d\to \mathbb{R}^{N^2}$ is a smooth, $d$-parameter family of square matrices.
On the other hand, I have not come across a nice expression for the second derivative (Hessian) of the determinant of such a family. Just by using Leibniz rule, one term is obvious: \operatorname{Tr}\left(\tilde{A}(s) A''(s)\right). However, I don't know of any nice expression of the derivative of the adjugate. Evaluating the adjugate component wise, we see that it should be linear in A'(s).
So the question is: does anyone know of a nice form to present the derivative of the Adjugate matrix, or a nice form to present the Hessian of the determinant?
For my purpose, an answer to the following weaker form of the problem would also be satisfactory:
Let $A:\mathbb{R}^d\to \mathbb{R}^{N^2}$ be a smooth $d$-parameter family of symmetric matrices. Suppose $\det(A(0)) = 0$. Now, if we know that the kernel of $A(0)$ as dimension at least 2, then by a diagonalisation argument, the adjugate matrix $\tilde{A}(0)$ must vanish identically. And therefore the Hessian is determined by the term with the derivative of the adjugate. Then under the assumptions that $d \geq 2$ and $A(0)$ has nullity at least 2: (a) is it possible for the Hessian to be non-degenerate? (b) If so, what are some sufficient conditions to guarantee nondegeneracy?
(In the case $d = 1$ it is clear that the "Hessian" can be non-degenerate, just by taking $N = 2$ and $A(s) = s I$.)