Show that the rate of change of the area of a circle with respect to its radius is equal to its circumference.

Solution The area $A$ = $A(r)$ of a circle of radius $r$ is $A(r) =\pi r^{2}$. The derivative function gives the rate of change of the area with respect to the radius. $$\begin{eqnarray*} A' ( r) &=&\lim\limits_{h\rightarrow 0}\dfrac{A( r+h) -A( r) }{h}=\lim\limits_{h\rightarrow 0}\dfrac{\pi ( r+h) ^{2}-\pi r^{2}}{h}\\[5pt] &=&\lim\limits_{h\rightarrow 0}\dfrac{\pi (r^{2}+2rh+h^{2})-\pi r^{2}}{h}=\lim\limits_{h\rightarrow 0}\dfrac{\pi h( 2r+h) }{h}\\[5pt] &=&\lim\limits_{h\rightarrow 0}\pi ( 2r+h) =2\pi r \end{eqnarray*}$$

The rate of change of the area of a circle with respect to its radius is the circumference of the circle, $2\pi r$.