Answer
$$\frac{{dV}}{{dr}} = 4\pi {r^2}$$
Work Step by Step
$$\eqalign{
& V = \frac{4}{3}\pi {r^3} \cr
& {\text{Using the definition of the derivative of a function}} \cr
& \frac{{dV}}{{dr}} = \mathop {\lim }\limits_{h \to 0} \frac{{V\left( {r + h} \right) - V\left( r \right)}}{h} \cr
& {\text{Then}} \cr
& \frac{{dV}}{{dr}} = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{4}{3}\pi {{\left( {r + h} \right)}^3} - \frac{4}{3}\pi {r^3}}}{h} \cr
& {\text{Simplifying}} \cr
& \frac{{dV}}{{dr}} = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{4}{3}\pi {{\left( {{r^3} + 3{r^2}h + 3r{h^2} + {h^3}} \right)}^3} - \frac{4}{3}\pi {r^3}}}{h} \cr
& \frac{{dV}}{{dr}} = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{4}{3}\pi {r^3} + 4\pi {r^2}h + 4\pi r{h^2} + \frac{4}{3}\pi {h^3} - \frac{4}{3}\pi {r^3}}}{h} \cr
& \frac{{dV}}{{dr}} = \mathop {\lim }\limits_{h \to 0} \frac{{4\pi {r^2}h + 4\pi r{h^2} + \frac{4}{3}\pi {h^3}}}{h} \cr
& \frac{{dV}}{{dr}} = \mathop {\lim }\limits_{h \to 0} \left( {4\pi {r^2} + 4\pi rh + \frac{4}{3}\pi {h^2}} \right) \cr
& {\text{Evaluating the limit}} \cr
& \frac{{dV}}{{dr}} = 4\pi {r^2} + 4\pi r\left( 0 \right) + \frac{4}{3}\pi {\left( 0 \right)^2} \cr
& \frac{{dV}}{{dr}} = 4\pi {r^2} \cr} $$
