Metric determinant and its partial and covariant derivative question : $\nabla_a \nabla_b \sqrt{g} \phi =\partial_a \sqrt{g} \partial_b \phi$ is true?
because $\nabla_a \sqrt{g}=0$ so we can write $\sqrt{g} \nabla_a \nabla_b \phi$ , but because the determinant of the metric does not transform like a scalar, we can not write partial derivative instead of covariant derivative.
 A: $$
g=\frac{1}{4!}\varepsilon^{abcd}\varepsilon^{efgh}g_{ae}g_{bf}g_{dg}g_{dh}\\
\therefore \quad \nabla_m g = \frac{1}{3!} \varepsilon^{abcd}\varepsilon^{efgh}g_{ae}g_{bf}g_{dg}\nabla_m g_{dh}\\=0\;.
$$ 
Note that $\varepsilon^{abcd}$ is Levi-Civita symbol, it is constant. It is tensor density of weight 0.5 and make $g$ be a tensor density of weight 1.
A: Yes it's true, since you can use the Leibniz rule like you do and since the covariant derivative of $\sqrt g$ is zero, like you say.
But your comment about partials and covariant derivatives is slightly misleading.  Whether it is a tensor or a tensor density is not the issue.  Unless the Christoffel symbols are zero at that point, you can never use partials instead of covariant derivatives.  And if they are zero, you can use partials no matter whether the tensor is a true tensor or a tensor density, since although the formula for the covariant derivative of a tensor density is a little different from that for a true tensor, it still involves just the partials and the Christoffel symbols.
