Let $M$ be a (pseudo-)Riemannian manifold with metric $g_{ab}$. Let $\nabla_a$ be the Levi-Civita connection on $M$. It's well-known that the Laplace—Beltrami operator can be defined in this context as $\nabla^2 = \nabla^a \nabla_a = g^{ab} \nabla_a \nabla_b$ where $g^{ab}$ is the dual metric and repeated indices are summed. However, we also have the coordinate formula $\nabla^2 \phi = \frac{1}{\sqrt{| \det g |}} \partial_a \left( \sqrt{| \det g |} g^{ab} \partial_b \phi \right)$ which, as I understand, comes from using the formula for the Hodge dual.
Without invoking advanced machinery, what is the easiest way to directly prove the equivalence of the two definitions? I can see that if the partial derivatives of $g_{ab}$ vanish, then $\nabla_{a} \left( g^{ab} \nabla_b \phi \right) = \partial_a \left( g^{ab} \partial_b \phi \right) = \frac{1}{\sqrt{| \det g |}} \partial_a \left( \sqrt{| \det g |} g^{ab} \partial_b \phi \right)$ and in the general case it suffices to prove that $\Gamma^{b}_{\phantom{b}ab} = \frac{1}{\sqrt{| \det g |}} \partial_a \left( \sqrt{| \det g |} \right)$ but then it is seems to be necessary to compute the derivative of a determinant. Is there a trick which can be used to avoid this calculation?