## Thomas' Calculus 13th Edition

$q'=\frac{2-2r}{3\sqrt[3] (2r-r^2)^2}$
Take the derivative and apply the chain rule: $q'=1/3(2r-r^2)^{-2/3}\times(2-2r)=\frac{2-2r}{3\sqrt[3] (2r-r^2)^2}$