 $$\beta\!\left(\partial_{\sqrt{j}}(f_0\star_B f_0)\right)=\beta\!\left(\partial_{\sqrt{j}}(f_0)\right)\cdot \beta\!\left(\partial_{\sqrt{j}}(f_0)\right)\ \text{modulo $\mathrm U(\mathfrak g)\cdot\mathfrak k^{-\delta+\frac{0}0\mathrm{tr}_\mathfrak g\circ\mathrm{ad}}$},$$ for $f_i$