Codifferential of volume form on Riemannian manifold.

Question

When I read some lecture note about forms on Riemannian manifold, there was a formula $$\delta(fw_g)=-i_{\nabla f}w_g$$ Where $f\in C^{\infty}(M^n)$, $w_g$ is the volume form of $M$. $\delta$ is the codifferential defined as $\delta:A^r(M)\rightarrow A^{r-1}(M), \delta=(-1)^{n(r-1)+1}\star d\star,$ and $i_{X}$ is the interior product w.r.t. vecotr $X$. What i tried to do is by using local O.N. frame ,then \begin{eqnarray*}\delta(fw_g)&=&-\sum i_{e_j}\nabla_{e_j}(fw_g)\\&=&-\sum i_{e_j}(e_j(f)w_g+f\nabla_{e_j}w_g)\\ &=& -\sum e_j(f)i_{e_j}w_g-\sum fi_{e_j}\nabla_{e_j}w_g\\ &=&-i_{\sum e_j(f)e_j}w_g-\sum fi_{e_j}\nabla_{e_j}w_g\\ &=&-i_{\nabla f}w_g-\sum fi_{e_j}\nabla_{e_j}w_g\end{eqnarray*} So I guess if all the deduction is correct, the second term should vanish, but I can't see how. Probably the fact that $w_g$ is volume form matter? Does it hold for any $p$-form?