We fix a Riemannian manifold $(M,g)$. Let $dV_g$ denote the Riemannian volume form which defines an isomorphism $* : C^\infty(M) \rightarrow \Omega^n(M)$ via $*f = fdV_g$. Another isomorphism that we have for a fixed $n-$form is given by $\beta : \mathfrak{X}(M) \rightarrow \Omega^{n-1}$ by contraction, i.e. $\beta(X) = \iota_XdV_g$.
Then we define the divergence operator $div : \mathfrak{X}(M) \rightarrow C^\infty(M)$ by $div(X) = *^{-1}d(\beta(X))$.
If $(x^1,\dots,x^n)$ is a set of coordinates for $M$ I want to show that $$div(X^i\partial_i) = \frac{1}{\sqrt{\det g}}\frac{\partial}{\partial x^i}(\sqrt{\det g}X^i).$$
The exercise is straightforward because its just using the definitions however I am lacking the denominator term. We have $\beta(X^i\partial_i) = \iota_{X^i\partial_i}dV_g = \sqrt{\det{g}}\iota_{X^i \partial_i}dx = \sqrt{\det{g}}X^idx_1 \wedge \dots \wedge \hat{dx^i} \wedge \dots \wedge dx^n$. Computing this differential yields $d\beta(X^i\partial_i) = \frac{\partial}{\partial x^i}(\sqrt{\det{g}}X^i)dx$. If we now use $*^{-1}$ which is just taking the function we get $$div(X^i\partial_i) = \frac{\partial}{\partial x^i}(\sqrt{\det{g}}X^i).$$
Notice that we could get the denominator if we differentiated the square root but then we would also get a $1/2$ term.
Thanks in advance!
Thank you @Kakashi for pointing out the detail. The detail is that $*f = fdV_g = f\sqrt{\det{g}}dx$. So that the function whose image under $*$ is $fdx$ is precisely $f/\sqrt{\det{g}}$.. $*(f/\sqrt{\det{g}}) = f/\sqrt{\det{g}}dV_g = fdx.$