A balloon, which always remains spherical,

The volume of a sphere (V) with radius (r) is given by,

$V=\frac{4}{3} \pi r^{3}$

It is given that:

Diameter $=\frac{3}{2}(2 x+1)$

$\Rightarrow r=\frac{3}{4}(2 x+1)$

$\therefore V=\frac{4}{3} \pi\left(\frac{3}{4}\right)^{3}(2 x+1)^{3}=\frac{9}{16} \pi(2 x+1)^{3}$

Hence, the rate of change of volume with respect to x is as

$\frac{d V}{d x}=\frac{9}{16} \pi \frac{d}{d x}(2 x+1)^{3}=\frac{9}{16} \pi \times 3(2 x+1)^{2} \times 2=\frac{27}{8} \pi(2 x+1)^{2}$