Derivatives of areas and volumes The derivative of the area $A(r) = \pi r^2$ of a circle with radius $r$ gives the circumfence $\frac{dA(r)}{dr} = 2\pi r$.
Similarly, the derivative of the volume $V(r) = \frac{4}{3}\pi r^3$ of the sphere (ball) with radius $r$ gives the surface area $\frac{dV(r)}{dr} = 4\pi r^2$.
This is not true say for the square or the square cuboid.
Just a coincidence. Stupid question I know.
 A: It still works with a cube, you just have to first define the radius of a cube. If we define the radius of the cube with side length $l$ to be $r=\frac12l$ (the shortest distance from the centre of the cube to the surface) then we get
$$V_{\text{cube}}=l^3=8r^3$$
$$A_{\text{cube}}=6l^2=24r^2=\frac{\mathrm{d}V_{\text{cube}}}{\mathrm{d}r}$$
More generally if we have a $3$ dimensional shape with some fixed length $l$ such that $V_{\text{shape}}=k_1l^3$ and $A_{\text{shape}}=k_2l^2$ then we can define $r=\frac{3k_1}{k_2}l$ such that the derivative of the volume with respect to $r$ gives the surface area.
A: The derivative  is  what it is for the circle (sphere) because you can compute the area (volume) as a sum of thin annuli (spherical shells). 
The corresponding integral has limits $0$ and $r$, where $r$ is the radius.
For the square (cube) the same argument with shells works, but the integral has limits $0$ and $s/2$, where $s$ is the side length. That explains the missing factor of $2$ when you differentiate the area (volume) to get the perimeter (area).
A: It is true for any shape which can be obtained as superposition of shells of same $d$-thickness, the thickness being measured normally to the surface.
So volumes which can be generated by expansion of parallel shells.
So it is clearly true for balls in whichever dimension.
Also (thanks to the comment by @Ethan Bolker) it is true for an equilateral triangle, since at the vertices the area not included by strips parallel to the edges is infinitesimal of higher order. Same for any regular polygon/polyhedron.
