Is it simply a coincidence that if you differentiate the formula for the volume of sphere you get the formula for the surface area of sphere?

Start with a sphere of radius $r$. Now let the radius of the sphere grow by some tiny amount $\Delta r$. How much has the volume changed? By the definition of the derivative, it has changed by approximately $$ \Delta r \cdot V'(r) $$ However, the added volume is basically a thin shell, and its volume is approximately equal to its surface area (the inner one, for convenience), multiplied by its thickness. This is $$ \Delta r\cdot SA(r) $$ Thus we have $$ \Delta r \cdot V'(r)\approx \Delta r\cdot SA(r)\\ V'(r)\approx SA(r) $$

Rigorous analysis of this setup will allow you to conclude that the approximation error above is small enough as $\Delta r$ becomes smaller, and thus that $V'(r)$ and $SA(r)$ are indeed equal.

How do you obtain the volume of a sphere? You just calculate a volume integral on the sphere

$$V=\int_{\mathbb{R}^3}\chi(x,y,z)\;d\mathbf{x}$$

where $\chi$ is a function that equals $1$ inside the sphere and $0$ outside. Of course is comfortable to switch to spherical coordinates. The determinat of the Jacobian is $|J|=r^2\sin\theta$, s0

$$ V=\int_0^Rr^2dr\int_0^\pi \sin\theta d\theta\int_0^{2\pi}d\varphi $$

calculating the two rightmost integral you obtain

$$ V=\int_0^R4\pi r^2dr =\int_0^R \frac{dV}{dr}dr \tag{1}$$

How do you calculate the surface area of a sphere? Through a surface integral

$$SA=\int_{\mathbb{R}^3}\sigma(x,y,z)\;d\mathbf{x}$$

where $\sigma(x,y,z)=\chi(x,y,z)\delta (r-R)$. In spherical coordinates:

$$ SA=\int_0^\pi r^2\sin\theta d\theta\int_0^{2\pi}d\varphi = 4\pi r^2 \tag{2}$$

So, confronting equation $(1)$ and $(2)$ is possible to prove that

$$SA=\frac{dV}{dr}$$

This, of course, means that

$dV=SAdr$

i.e., the infinitesimal increment of the Volume $dV$ is obtained through the product of the surface area $SA$ and the infinitesimal increment of the radius $dr$.