Find the derivative of the following function (it is to be understood that $a, b, c, d, p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\frac{ax+b}{px^{2}+qx+r}$

Let $f(x) = \frac{ax+b}{px^{2}+qx+r}$.
Using the quotient rule,$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^{2}}$:
$f'(x) = \frac{(px^{2}+qx+r) \frac{d}{dx}(ax+b) - (ax+b) \frac{d}{dx}(px^{2}+qx+r)}{(px^{2}+qx+r)^{2}}$
$f'(x) = \frac{(px^{2}+qx+r)(a) - (ax+b)(2px+q)}{(px^{2}+qx+r)^{2}}$
$f'(x) = \frac{apx^{2} + aqx + ar - (2apx^{2} + aqx + 2bpx + bq)}{(px^{2}+qx+r)^{2}}$
$f'(x) = \frac{apx^{2} + aqx + ar - 2apx^{2} - aqx - 2bpx - bq}{(px^{2}+qx+r)^{2}}$
$f'(x) = \frac{-apx^{2} - 2bpx + ar - bq}{(px^{2}+qx+r)^{2}}$