Discrete product rule for $D_x^c(u\bar{u})$

For $(u_i)_{i\in I}$ define the central gradient by$$(D_x^c u)_i = \frac{u_{i+1}-u_{i-1}}{2\Delta x}\textrm{ for all }i\in I.$$

If, additionally, I have some $(\bar{u}_i)_{i\in I}$, I am wondering what$$(D_x^c(u\bar{u}))_i$$looks like and if there is some kind of a product rule version.

If we define$$(D_x^+ u)_i:=\frac{u_{i+1}-u_i}{\Delta x},\quad (D_x^- u)_i:=\frac{u_i-u_{i-1}}{\Delta x},$$then $D_x^c = \frac{1}{2}(D_x^++D_x^-)$.

Now, for $\Delta_i u_i=u_{i+1}-u_i$, there is the product rule$$\Delta_i(u\bar{u})_i=u_{i+1}\Delta_i\bar{u}_i+\Delta_i u_i\bar{u}_i,$$see, for instance, here.

Writing $(D_x^cu)_i=\frac{1}{2\Delta x}(\Delta_i u_i-\Delta_{i-1}u_{i-1})$, one should get$$\begin{align*}(D_x^c (u\bar{u}))_i&=\frac{1}{2\Delta x}(\Delta_i(u\bar{u})_i+\Delta_{i-1}(u\bar{u})_{i-1})\\&=\frac{1}{2\Delta x}(u_{i+1}\Delta_i\bar{u}_i+\Delta_i u_i\bar{u}_i+u_i\Delta_{i-1}\bar{u}_{i-1}+\Delta_{i-1} u_{i-1}\bar{u}_{i-1})\\&=\frac{1}{2\Delta x}(u_{i+1}(\bar{u}_{i+1}-\bar{u}_i)+(u_{i+1}-u_i)\bar{u}_i+u_i(\bar{u}_i-\bar{u}_{i-1})+(u_i-u_{i-1})\bar{u}_{i-1})\\&=\frac{1}{2}(u_{i+1} (D_x^+\bar{u})_i+\bar{u}_i (D_x^+ u)_i+u_i (D_x^-\bar{u})_i+\bar{u}_{i-1}(D_x^- u)_i)\end{align*}$$Can this be further simplyfied in order to have only $(D_x^c u)_i$ and $(D_x^c\bar{u})_i$ appearing on the right-hand side so that a "product rule" is more obvious?