Why isn't $ SA_{\text{cube}} = 3x^2 $?

Solution 1:

If you express the length of the cube in terms of its half-length $y=x/2$, then the volume is $V=8y^3$ and the surface area is $S = 24y^2$, so you have $S=\mathrm{d}V/\mathrm{d}y$ just like for the sphere.

Solution 2:

If you compare a cube of side length $x$ to a cube of side length $x + \epsilon$, the difference is a bunch of slices corresponding to the faces, but they have width $\frac{\epsilon}{2}$, not $\epsilon$. As Chris Taylor says, to get the expected normalization you work in terms of half of the side length. Then you get

$$(2y + 2 \epsilon)^3 = 8(y^3 + \epsilon (3y^2) + \epsilon^2 (3y) + \epsilon^3)$$

where the Taylor coefficients are, in order: the volume $8y^3$, the surface area $24y^2$ times $\epsilon$, the total length $24y$ of the edges times $\epsilon^2$, and the number $8$ of vertices times $\epsilon^3$.