Answer
$16\pi$
Work Step by Step
Let us consider the function of the volume of the ball with a radius $r$ be $f(r)=(\dfrac{4}{3})\pi r^3$
The rate of change of the volume $V$ of the ball with respect to $r=2$ is defined as the derivative $f'(2)$
This implies that
$f'(2)=\lim_{h\to 0}\dfrac{f(h+2)-f(2)}{h}=\lim\limits_{h\to 0}\dfrac{\dfrac{4}{3}\pi(h+2)^3-\dfrac{4}{3}\pi \times2^3}{h}=\lim\limits_{h\to 0}(\dfrac{4}{3}\pi h^2+8\pi h+16\pi)$
Hence, $f'(2)=\dfrac{4}{3}\pi(0)^2+8\pi(0)+16\pi=16\pi$
