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I'm talking about these specific relations, where g is the determinant of the metric tensor (so it's symmetric spscific), which is a function of $x^k$:

$\frac{1}{2g}\frac{\partial g}{\partial x^k}=\frac{1}{2}\frac{\partial \ln g}{\partial x^k}$

I've tried making discrete variations of the matrix, but I can't reach this particular relation.

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up vote 1 down vote accepted

This follows from chain rule. In general, if you want some identities involving derivatives of matrices, I would recommend you to look at "The Matrix Cookbook"

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Holy heaping helpings of high fructose corn syrup! This is so obvious I'm actually embarassed. I just got caught up in the fact that I'm working with matrices and I logarithm of matrices is a little different that I forgot it's determinant is just a simple function. – Diego Feb 20 '11 at 4:43

This formula is a direct consequence of the chain rule.

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