This question explains why the derivative of the volume of a sphere is equal to its surface area. This is based on the logic using the differential.
However, to what extent (and when can this be generalized). For example
A. The derivative of the volume of a cube with side s does not equal its surface area.
B. The derivative of the volume of a cube with side 2r does equal its surface area.
What is going on here under the hood? Can anyone explain without using Maths beyond Calc I/II.
Also, I am not so interested in understanding why the derivative of the area is equal to the circumfence, as in understanding which shapes in general exhibit this property and why.
Note: This paper explains this question, but is quite technical. Summarizing the main results in less technical language would be useful.
When built upon a single symmetric length parameter L (sphere radius, side of cube or any Platonic solid radius icosahedron &c.,) we could generally say:
$$ V=aL^3;\; A= b L^2;\;\dfrac{V^2}{A^3}= \dfrac{a^2}{b^3} = \text{constant} \tag 1 $$
By logarithmic differentiation
$$ \dfrac{2\; dV}{V}=\dfrac{3\; dA}{A}\;$$
$$\dfrac{dV}{dA}=\dfrac{3 V}{2 A} \tag2, $$
a general relation like for a cube, sphere or dodecahedron, ( that always works out in terms of their characteristic lengths $L =(a/2, r/2, R/2)$ respectively.)
EDIT1:
Can be differentiated also this way $$ V= a L^3,\dfrac{dV}{dL}=3 a L^2 $$
$$ A= b L^2, \dfrac{dA}{dL}=2 b L $$
Divide directly $$\dfrac{V}{A}=\dfrac{ A L }{b} \tag 3 $$ Divide differentials
$$\dfrac{dV}{dA}=\dfrac{3 A L }{2b} \tag 4 $$
From (3) and (4)
$$\dfrac{dV}{dA}=\dfrac{3 V}{2 A} \tag {5=2}. $$