#### Answer

$\frac{dq}{dr} = \frac{2(1-r)}{3(2r-r^{2})^{\frac{2}{3}}}$

#### Work Step by Step

$q = \sqrt[3] u$
where,
$u = (2r-r^{2})$
Now,
$\frac{dq}{du} = \frac{1}{3u^{\frac{2}{3}}} $
and,
$\frac{du}{dr} =2-2r= 2(1-r)$
So,
$\frac{dq}{dr} = \frac{dq}{du} \times \frac{du}{dr}$
$\frac{dq}{dr} = \frac{2(1-r)}{3u^{\frac{2}{3}}}$
$\frac{dq}{dr} = \frac{2(1-r)}{3(2r-r^{2})^{\frac{2}{3}}}$