Derivative of the determinant ($D(\det {)}_A$)

For an invertible matrix $A$ and any matrix $H$,Indeed $\det (A+tH)=\det (A)\det (I+t{A}^{-1}H)$, and multilinearity shows that the linear coefficient of $\det (I+tB)$ is $\mathrm{tr}B$. At a singular matrix the derivative is instead written $\mathrm{tr}(\mathrm{adj}(A)H)$; this polynomial formula holds for all $A$.

For an arbitrary square matrix $A$, the derivative of the determinant in direction $H$ is the displayed expression. Differentiating the determinant's multilinear expansion gives the sum of entries of $H$ weighted by their cofactors, equal to the displayed trace. If $A$ is invertible, $\mathrm{adj}(A)=(\det A)A^{-1}$ gives $(\det A)\mathrm{tr}(A^{-1}H)$, and division by $\det A$ gives the Jacobi determinant derivative formula for the logarithmic derivative. The adjugate expression remains valid at singular matrices.

The derivative of the determinant at an invertible matrix is $D{\det }_A(K)=\det A\mathrm{tr}(A^{-1}K)$. At $A=I+H$, use $\det (I+H)=1+\mathrm{tr}H+O(\Vert H{\Vert }^2)$ and $(I+H{)}^{-1}=I-H+O(\Vert H{\Vert }^2)$. Differentiating the first derivative gives the displayed symmetric bilinear map. More generally $D^2{\det }_A(H,K)=\det A[\mathrm{tr}(A^{-1}H)\mathrm{tr}(A^{-1}K)-\mathrm{tr}(A^{-1}H{A}^{-1}K)]$.