A balloon, which always remains spherical, has a variable diameter $\cfrac{3}{2}(2x + 1)$. Find the rate of change of its volume with respect to $x$.

Diameter of the balloon, $d = \cfrac{3}{2}(2x + 1)$

Therefore the radius of the balloon, $r = \cfrac{d}{2} = \cfrac{1}{2}\left\{ {\cfrac{3}{2}(2x + 1)} \right\} = \cfrac{3}{4}(2x + 1)$

So, the volume $V$ of the balloon
$V = \cfrac{4}{3}\pi {\left( {{\rm{radius}}} \right)^3} = \cfrac{4}{3}\pi {\left\{ {\cfrac{3}{4}\left( {2x + 1} \right)} \right\}^3} = \cfrac{{9\pi }}{{16}}{(2x + 1)^3}$ …(i)

Differentiating (i) w.r.t. $x$, we get
$\cfrac{{dV}}{{dx}} = \cfrac{{9\pi }}{{16}} \times 3{(2x + 1)^2} \times 2 = \cfrac{{27\pi }}{8}{(2x + 1)^2}$