As the title says, is it just a coincidence that $d(ax^2+bx+c)/dx=\pm \sqrt{\Delta}$? (where $\Delta=b^2-4ac$, i.e. discriminant of the quadratic). We can get this easily from rearranging the quadratic formula:
$$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
$$\iff 2ax+b=\pm \sqrt{b^2-4ac}$$
$$\iff \frac{d(ax^2+bx+c)}{dx}=\pm \sqrt{\Delta}$$
but that doesn't explain why it's true, i.e. why the derivative of a quadratic equals $\pm$ square root of $\Delta$. Seems a touch mysterious.
EDIT: I found this in a book entitled "Vedic Mathematics" by Bharati.
Writing the function with its roots seems helpful. The derivative can be written with terms like $x_1-x_2$ which its product is related to the discriminant.
Let the roots of $f_2(x)=ax^2+bx+c=0$ be $x_1$ and $x_2$. Then $f_2(x)=a(x-x_1)(x-x_2)$ and $\Delta=(a(x_1-x_2))^2$. Differentiation gives $f'_2(x_1)=a(x_1-x_2)=\sqrt{\Delta}$.
Looking at the cubic function, $f_3(x)=a(x-x_1)(x-x_2)(x-x_3)$, we have $f'_3(x_1)=a(x_1-x_2)(x_1-x_3)$. And $\Delta=a^4((x_1-x_2)(x_2-x_3)(x_3-x_1))^2$. So similar result does not hold because $(x_2-x_3)$ is missing. But their product $f'_3(x_1)f'_3(x_2)f'_3(x_3)=\Delta/a$ and in this way, the equation can be generalized to order $n$ polynomials as $$\prod_{k=1}^nf'(x_k)=\left(\frac{\Delta}{a^{n-2}}\right)^{(n-1)/2}$$