Differentiation of the determinant $g$ Let $g$ be the determinant of the metric tensor.
I want to derive the following equation $g_{,\nu}=gg^{\lambda \mu}g_{\lambda \mu,\nu}$. It is said that $gg^{\lambda \mu}$ is a cofactor, but I can't understand why. To begin with, I'm not familiar with how to express the determinant of the metric tensor i.e. $g$. I know that $g$, the determinant is the sum of the cofactors multiplied with corresponding matrix element.
Can I get some illuminations on this?
 A: Determinant of the metric, or any matrix can be expressed through Levi-Civita (relative) tensors, or through generalized Kroenecker deltas
I have written about it here
https://www.physicsforums.com/threads/i-cant-verify-a-relationship-between-cofactor-and-determinant.970419/post-6165630
A: Let $A$ be a $n\times n$ matrix.
Recall that $$A \text{adj}(A)=\det(A)\cdot1$$ where $1$ is the identity matrix, $\text{adj}(A)=C^T$ is the adjoint matrix, $C_{ij}=(-1)^{i+j}M_{ij}$ is the cofactor matrix, $M_{ij}$ is the minor of the $A_{ij}$. This is justified because $$\det{A}=\sum_j A_{ij}C_{ij}=\sum_j A_{ij}adj_{ji}=(A\text{adj}(A))_{ii}.$$
Now consider
$$d(\det(A))=\sum_{i,j}\frac{\partial\det(A)}{\partial A_{ij}}dA_{ij}$$
where
$$\frac{\partial\det(A)}{\partial A_{ij}}=\frac{\partial\sum_k A_{ik} \text{adj}_{ki}}{\partial A_{ij}}=\sum_k\frac{\partial A_{ik}\text{adj}_{ki}}{\partial A_{ij}}=\sum_k \frac{\partial A_{ik}}{\partial A_{ij}}\text{adj}_{ki}+A_{ik}\frac{\partial \text{adj}_{ki}}{\partial A_{ij}}=\sum_k \delta_{jk}\text{adj}_{ki}=\text{adj}_{ji}$$
since $\text{adj}_{ji}(A)=C_{ij}$ is independent of $A_{ij}$ as it is the determinant of the matrix A with $i^{th}$ row and $j^{th}$ column removed.
Thus we have
$$d(\det(A))=\sum_{i,j}\text{adj}_{ji} dA_{ij}=\sum_{i,j} \text{adj}_{ij} A_{ji} = tr(\text{adj(A)} dA)=tr(\det(A)A^{-1}dA).$$
Finally putting $(A)=(g)$, we conclude
$$g_{,\nu}=gg^{\mu\lambda}g_{\mu\lambda ,\nu}.$$
