At $v \ll c$, the Taylor expansion gives $1/\sqrt{1-v^2/c^2} \approx 1 + \tfrac12 v^2/c^2$. Comparing the classical kinetic-energy expression $\tfrac12(\Delta m)v^2$ for the lost mass $\Delta m$ with the right side, the coefficients of $v^2$ on both sides must match.