$\mathrm{\nabla}\cdot \overrightarrow{B}=0\phantom{\rule{thinmathspace}{0ex}}.$

The derivation in Griffiths Introduction to elecrodynamics uses

$\mathrm{\nabla}\cdot \overrightarrow{B}\text{}=\text{}\frac{{\mu}_{0}}{4\pi}\int \mathrm{\nabla}\cdot (\overrightarrow{J}\times \frac{\overrightarrow{r}}{{r}^{3}})\phantom{\rule{thinmathspace}{0ex}},$

but to use this, I would need $\overrightarrow{J}\times \frac{\overrightarrow{r}}{{r}^{3}}$ and $\mathrm{\nabla}\cdot (\overrightarrow{J}\times \frac{\overrightarrow{r}}{{r}^{3}})$ to be continuous.

Since $\overrightarrow{J}=\rho \overrightarrow{v}$ , then if $\rho $ (charge density) has a continuous derivative, I'm done; but, I'm not sure why can I affirm this because clearly, I can construct charge distributions with jump discontinuities.