When discussing integration in curvilinear coordinate systems, the determinant often appears as a factor in the "volume element." In General Relativity, the action of the Einstein field equations contains the determinant of the metric, and when performing variations on it, one naturally encounters the problem of differentiating a determinant. I have referred to Landau’s The Classical Theory of Fields and Foundations of Mathematical Physics: An Introduction to Linear Higher Mathematics for Physics to understand the relevant results, which I have recorded as follows.
Derivation
Let \boldsymbol{A}(t)=\left(a_{ij}(t)\right)_{n\times n} be an n-th order matrix where each element is a function of t. Its determinant is denoted by |\boldsymbol{A}|. Naturally, we consider \frac{d}{dt}|\boldsymbol{A}| From the fundamental properties of determinants, we have |\boldsymbol{A}(a_{ij}+\varepsilon)|-|\boldsymbol{A}|=\varepsilon A_{ij} where |\boldsymbol{A}(a_{ij}+\varepsilon)| is the value of the determinant after replacing the element a_{ij} of matrix \boldsymbol{A} with a_{ij}+\varepsilon, and A_{ij} is the cofactor of the determinant |\boldsymbol{A}| with respect to a_{ij}. The above equation gives \frac{\partial |\boldsymbol{A}|}{\partial a_{ij}}= A_{ij} In other words, the cofactor can be expressed as the partial derivative of the determinant.
Then \frac{d}{dt}|\boldsymbol{A}|=\sum_{i}\sum_{j}\frac{\partial |\boldsymbol{A}|}{\partial a_{ij}}\frac{d a_{ij}}{dt}=\sum_{i}\sum_{j} A_{ij}\frac{d a_{ij}}{dt} (To derive the first equality, one only needs to add an infinitesimal amount to every element of matrix \boldsymbol{A}, expand the determinant of the incremented matrix, and retain the first-order infinitesimal terms.)
Therefore, \frac{d}{dt}|\boldsymbol{A}|=|\boldsymbol{A}|\sum_{i}\sum_{j} \frac{A_{ij}}{|\boldsymbol{A}|}\frac{d a_{ij}}{dt} Perhaps rewriting this will make the process clearer: \frac{d}{dt}|\boldsymbol{A}|=|\boldsymbol{A}|\sum_{j=k}\sum_{i} \frac{A_{ij}}{|\boldsymbol{A}|}\frac{d a_{ik}}{dt} where \frac{A_{ji}}{|\boldsymbol{A}|} is exactly the (i,j)-th element (\boldsymbol{A}^{-1})_{ij} of the inverse matrix \boldsymbol{A}^{-1}, and \frac{d a_{ij}}{dt} is an element of the matrix \frac{d \boldsymbol{A}}{dt}. The first summation corresponds to the multiplication of two matrices: \left(\boldsymbol{A}^{-1}\frac{d \boldsymbol{A}}{dt}\right)_{jk}=\sum_{i} (\boldsymbol{A}^{-1})_{ji}\left(\frac{d\boldsymbol{A}}{dt}\right)_{ik}=\sum_{i} \frac{A_{ij}}{|\boldsymbol{A}|}\frac{d a_{ik}}{dt} The second summation is equivalent to taking the trace of the matrix, so \frac{d}{dt}|\boldsymbol{A}|=|\boldsymbol{A}| \text{Tr}\left(\boldsymbol{A}^{-1}\frac{d \boldsymbol{A}}{dt}\right) Using notation from tensor analysis, this becomes even simpler. Let g=\det(g_{\mu\nu}) and g_{\mu s} g^{s\nu}=\delta_{\mu}^{\nu}, then dg=g g^{\mu\nu} dg_{\mu\nu}
Example
As a simple application, let us find \frac{d}{dt}|\boldsymbol{I}+t\boldsymbol{A}| where \boldsymbol{I} is the identity matrix and \boldsymbol{A} is a square matrix of the same size. Substituting into the formula, we get \frac{d}{dt}|\boldsymbol{I}+t\boldsymbol{A}|=|\boldsymbol{I}+t\boldsymbol{A}|\text{Tr}\big[(\boldsymbol{I}+t\boldsymbol{A})^{-1}\boldsymbol{A}\big] Then \left(\frac{d}{dt}|\boldsymbol{I}+t\boldsymbol{A}|\right)_{t=0}=\text{Tr}\,\boldsymbol{A} According to the Taylor expansion, when |t| is sufficiently small, we have the approximation: |\boldsymbol{I}+t\boldsymbol{A}|\approx 1 + t\times\text{Tr}\,\boldsymbol{A}
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