areacalculuscirclesderivativesgeometry
When differentiated with respect to $r$, the derivative of $\pi r^2$ is $2 \pi r$, which is the circumference of a circle.
Similarly, when the formula for a sphere's volume $\frac{4}{3} \pi r^3$ is differentiated with respect to $r$, we get $4 \pi r^2$.
Is this just a coincidence, or is there some deep explanation for why we should expect this?
Actually, it is also true for squares (and for regular polygons in general!). The problem you ran into is what the equivalent of "r" is. The side length of a square is actually more comparable to the circle's diameter.
Instead, the correct analogue of the circle's radius is the distance from the center of the square to the midpoint of one side, which is only half as long as the square's side.
Here, we have $A = (2r)^2 = 4 r^2$ and $P = 4 (2r) = 8 r$. The perimeter is the derivative of the area with respect to $r$, just as in the case of a circle.
What you need to look at is the second theorem of Pappus. When you rotate things in a circle, you have to account for the difference in the distance traveled by the point furthest from the axis (the bottom corner of our triangle) which goes the full $2\pi$ around. However, the vertical side doesn't actually move, so it doesn't get the full $2\pi$.
Pappus' theorem says you can use the radius of the centroid as your revolution radius, and it just so happens the the centroid of this triangle is a distance $r/3$ from the axis.
Another good read is this.
Best Answer
Consider increasing the radius of a circle by an infinitesimally small amount, $dr$. This increases the area by an annulus (or ring) with inner radius $2 \pi r$ and outer radius $2\pi(r+dr)$. As this ring is extremely thin, we can imagine cutting the ring and then flattening it out to form a rectangle with width $2\pi r$ and height $dr$ (the side of length $2\pi(r+dr)$ is close enough to $2\pi r$ that we can ignore that). So the area gain is $2\pi r\cdot dr$ and to determine the rate of change with respect to $r$, we divide by $dr$ and so we get $2\pi r$. Please note that this is just an informative, intuitive explanation as opposed to a formal proof. The same reasoning works with a sphere, we just flatten it out to a rectangular prism instead.