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The determinant of a matrix $A(t)=A_{ij}(t)$ is given by $$det(A_{ij}(t))=e_{ijk}A_{i1}(t)A_{j2}(t)A_{k3}(t),$$ where $e_{ijk}$ is the Cartesian alternator. Let us take the derivative of this expression with respect to the parameter $t$, $$\frac{d}{dt}det(A_{ij}(t))=e_{ijk}(A_{i1}(t),tA_{j2}(t)A_{k3}(t)+A_{i1}(t)A_{j2}(t),tA_{k3}(t)+A_{i1}(t)A_{j2}(t)A_{k3}(t),t),$$ where $,t$'' stands for derivation. Let us now consider the quantity $$C_{k3}(t)=e_{ijk}A_{i1}(t)A_{j2}(t)=\frac{1}{2}(e_{ijk}A_{i1}(t)A_{j2}(t)+e_{jik}A_{j1}(t)A_{i2}(t))=\frac{1}{2}e_{ijk}(A_{i1}(t)A_{j2}(t)-A_{i2}(t)A_{j1}(t)),$$ which is the same as $$C_{k3}(t)=\frac{1}{2}e_{ijk}e_{lm3}A_{il}(t)A_{jm}(t),$$ and in general, $$C_{kn}(t)=\frac{1}{2}e_{ijk}e_{lmn}A_{il}(t)A_{jm}(t).$$ This is the so-called adjugate'' $adj$ of the matrix $A(t)$ (notice that this formula is also useful to find the inverse of $A(t)$ whenever it be no singular). Thus, the sum for $\frac{d}{dt}det(A_{ij}(t))$ above, may be expressed by $$\frac{d}{dt}det(A_{ij}(t))=\frac{1}{2}e_{ijk}e_{lmn}A_{il}(t)A_{jm}(t)A_{kn}(t),t=C_{kn}(t)A_{kn}(t),t,$$ which is clearly the trace'' $tr$ of this product. So that, in matrix notation we have $$\frac{d(detA(t))}{dt}=tr(adj(A(t))\frac{d(A(t))}{dt}).$$ Indeed this formula is also valid for any finite dimension $n$, and its proof is a little more elaborated. We have exposed here the case for $n=3$ for reason of brevity. Also note that $C_{kn}^T=C_{nk}$ is a cofactor of the matrix $A(t).$