We know the area of the circle is $$A(r)=\pi r^2$$Differentiate it with respect to $r$. $$A'(r)=2\pi r$$ which turns out to be the circumference of the circle.
A similar behaviour is observed with the volume of sphere $$V(r)=\frac{4}{3}\pi r^3$$ Differentiate with respect to $r$.$$V'(r)=4\pi r^2$$ which coincidently turns out to be the surface area of the volume.
Is this result purely coincidental?
The property is also found in cubes (in some fashion)
$$S(r) = 6r^2 = 2\frac{dr^3}{dr} = 2V'(r)$$ Is there something special about these shapes which can be modeled by the function$$V'(r)= kS(r)$$
Note that area of the circle $A$ and volume $V$ of the sphere can be obtained as
$A(r)=2\pi\int_0^r xdx$
$V(r)=4\pi\int_0^r x^2dx$