Why is the derivative of circle area the circumference? Why is the derivative of the volume of a sphere the surface area? And why is the derivative of the area of a circle the circumference? Too much of a coincidence, there has to be a reason! 
Also, why is the sum of the cubes the square of the sum of the integers?

I'll answer the last question first because that's the easiest one in my opinion.

Prove that
$$1^3 + 2^3 + 3^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$$
We will prove this using induction, by first confirming the identity when $n = 1$ and then proving the inductive step.
Base Case: $n = 1$
$$1^3 = \left(\frac{1(1+1)}{2}\right)^2 = \left(\frac{2}{2}\right)^2 = 1^2 = 1$$
Inductive Step:
Suppose the equality is true for some $x$. That is, $$1^3 + 2^3 + 3^3 + \cdots + x^3 = \left(\frac{x(x+1)}{2}\right)^2$$
Now we add $(x+1)^3$ to both sides and get
$$1^3 + 2^3 + 3^3 + \cdots + x^3 + (x+1)^3 = \left(\frac{x(x+1)}{2}\right)^2 + (x+1)^3 = (x+1)^2 \left(\frac{x^2}{4} + x + 1\right) = \frac{(x+1)^2}{4} (x^2 + 4x + 4) = \frac{(x+1)^2 (x+2)^2}{4} = \left(\frac{(x+1)(x+2)}{2}\right)^2$$
Therefore, if the equation is true for some $x$, then it is also true for $x+1$. And because it is true for $1$, it will be true for $2$, for $3$, for $4$, etc.

Prove that the derivative of the area of the circle is the circumference.

We note that the area of the bigger circle is $\pi (r+dr)^2$ and that of the smaller circle is $\pi r^2$. The difference in area, therefore, is $\pi (r+dr)^2 - \pi r^2 = \pi (2rdr + dr^2)$
We can then imagine turning the small "outline" (the area between the large circle and the small circle) into a thin, narrow strip. Assuming that $dr$ is small enough, the strip will be almost a rectangle with area $\pi (2rdr + dr^2)$. Because the height of the rectangle is $dr$, we divide the area by the height to get the width, which is $$\frac{\pi (2rdr + dr^2)}{dr} = \pi (2r + dr)$$
And finally, because $dr$ approaches $0$ as the strip gets thinner, we see that the width approaches $2\pi r$, the circumference.
Similar logic applies to the sphere and the surface area.
