Find the derivative of the following function (it is to be understood that $a, b, c$ are fixed non-zero constants): $\frac{1}{ax^{2}+bx+c}$

Let $f(x) = \frac{1}{ax^{2}+bx+c}$.
Using the quotient rule,where $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^{2}}$:
$f^{\prime}(x) = \frac{(ax^{2}+bx+c) \frac{d}{dx}(1) - (1) \frac{d}{dx}(ax^{2}+bx+c)}{(ax^{2}+bx+c)^{2}}$
Since $\frac{d}{dx}(1) = 0$ and $\frac{d}{dx}(ax^{2}+bx+c) = 2ax+b$:
$f^{\prime}(x) = \frac{(ax^{2}+bx+c)(0) - (2ax+b)}{(ax^{2}+bx+c)^{2}}$
$f^{\prime}(x) = \frac{-(2ax+b)}{(ax^{2}+bx+c)^{2}}$