Let $g_{ij}$ be a Riemannian metric of a Riemannian manifold, prove:
$$ g^{ij}\frac{\partial g_{ij}}{\partial x^k}=\frac{1}{G}\frac{\partial G}{\partial x^k},\quad G=\det(g_{ij}),\quad g^{ij}=g_{ij}^{-1}$$ I have tested the case $i,j\leq 2$, it's correct, but in the general case, I think I lack the suitable skill to prove it. So can someone help me?
The derivative of the determinant is given by Jacobi's formula: $$ \partial_k \det(g) = \det(g)\,\text{tr}\left( g^{-1}\partial_kg \right) $$ This simplifies nicely in indicial form: \begin{align} \partial_k \det(g) &= \det(g)g^{\alpha\beta}\partial_kg_{\beta\gamma}\delta_\alpha^\gamma\\ &= \det(g)g^{\alpha\beta}\partial_kg_{\beta\alpha}\\ &= \det(g)g^{ij}\partial_kg_{ij}\\ \end{align} Therefore, we get: $$ \frac{1}{G}\frac{\partial G}{\partial x^k}= \frac{1}{\det(g)}\partial_k\det({g}) = g^{ij}\partial_kg_{ij} = g^{ij}\frac{\partial g_{ij}}{\partial x^k} $$