general relativity - 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 metric determinant is not transfers like scalar, we can not write partial derivative instead of covariant derivative. what do think guys ?