Answer
The rate of change of the volume $V$ of the ball with respect to $r=2$ is $16\pi$.
Work Step by Step
The function of the volume of the ball with a radius $r$ is $f(r)=(4/3)\pi r^3$.
The rate of change of the volume $V$ of the ball with respect to $r=2$ is the derivative $f'(2)$, which can be calculated by $$f'(2)=\lim_{h\to0}\frac{f(h+2)-f(2)}{h}=\lim_{h\to0}\frac{\frac{4}{3}\pi(h+2)^3-\frac{4}{3}\pi \times2^3}{h}$$
$$f'(2)=\lim_{h\to0}\frac{\frac{4}{3}\pi(h^3+6h^2+12h+8)-\frac{4}{3}\pi\times8}{h}$$
$$f'(2)=\lim_{h\to0}\frac{\frac{4}{3}\pi h^3+8\pi h^2+16\pi h}{h}=\lim_{h\to0}\Big(\frac{4}{3}\pi h^2+8\pi h+16\pi\Big)$$
$$f'(2)=\frac{4}{3}\pi\times0^2+8\pi\times0+16\pi=16\pi$$
The rate of change of the volume $V$ of the ball with respect to $r=2$ is $16\pi$.
