In B. Schutz's book A First Course in General Relativity, I read on the page 152:
$$ \Gamma^\alpha{}_{\mu \alpha} = \frac{1}{2} g^{\alpha \beta} g_{\alpha \beta, \mu} \tag{6.38} $$ Since $\left(g^{\alpha \beta}\right)$ is the inverse matrix of $\left(g_{\alpha \beta}\right)$, it can be shown (see Exer. 7, $\S 6.9$ ) that the derivative of the determinant $g$ of the matrix $\left(g_{\alpha \beta}\right)$ is: $$ g_{, \mu}=g g^{\alpha \beta} g_{\beta \alpha, \mu} \tag{6.39} $$ Using this in Eq. (6.38), we find: $$ \Gamma^\alpha{}_{\mu \alpha}=(\sqrt{-g})_{, \mu} / \sqrt{-g} \tag{6.40}$$
Where does the square root come from in the Eq. 6.40?
Using Eq. (6.39) in Eq. (6.38), we just find:
$$ \Gamma^\alpha{}_{\mu \alpha}=\frac{1}{2} g_{, \mu} / g $$
Perhaps I am missing some obvious facts. Nevertheless, if someone would be so kind to explain to me, I would appreciate greatly.
