A balloon, which always remains spherical, has a variable diameter $\dfrac{3}{2} \left(2x+1\right)$. Find the rate of change of its volume with respect to x.
Let the diameter of the sphere $=D = \dfrac{3}{2} \left(2x+1\right)$
Volume of sphere $V = \dfrac{4}{3}\pi \times \left(\dfrac{D}{2}\right)^3$
i.e. $V = \dfrac{4\pi}{24} \times D^3$
i.e. $V = \dfrac{\pi}{6} \times \left[\dfrac{3}{2} \left(2x+1\right)\right]^3$
i.e. $V = \dfrac{27\pi}{48}\left(8x^3+12x^2+6x+1\right)$
i.e. $V = \dfrac{9\pi}{16} \left(8x^3+12x^2+6x+1\right)$ $\;\; \cdots$ (1)
Differentiating equation (1) w.r.t x gives
$\dfrac{dV}{dx}= \dfrac{9\pi}{16}\left(24x^2+24x+6\right)$
i.e. $\dfrac{dV}{dx}= \dfrac{27\pi}{8}\left(4x^2+4x+1\right)$
i.e. $\dfrac{dV}{dx} = \dfrac{27 \pi}{8}\left(2x+1\right)^2$