Question: So my question is this:
$$V=\frac{4}{3}\pi r^3$$ 
And, 
$$\frac{dV}{dr}=4\pi r^2=SA$$
Is this a coincidence or are there some mathematical hoodoos that I'm unaware of?

P.S. are there any more tags that I should use?

Answer: 
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.
I think I have something but I'm not sure what. 

The volume of an object is the integral $\int 1 d\tau$ where $\tau$ is the volume element and the boundaries are the boundary of the surface. 

The divergence of $\vec{r}/3$ is 1 in any co-ordinate system. So the above integral can be expressed as the volume integral of a divergence. Where $\vec{r}$ is the position vector. 

By Gauss' Law $\int \nabla\cdot (\vec{r}/3) d\tau=\int \frac{\vec{r}}{3}\cdot \hat{n} dA$ 

Where $\hat{n}dA$ is the vector form of the infinitesimal surface area. 

For a figure of constant radius, $\hat{n}=\hat{r}$, so the integrand on the right becomes (r/3). 

I'm not sure what comes next, but there does seem to be a tie in there between the entire volume and only considering geometric features of the boundary.