Question

Find the derivative of $\frac{p{x}^2+qx+r}{ax+b}$ where $p,q,r,a,b$ are fixed non-zero constants.

Answer 1

Let $f\left(x\right)=\frac{p{x}^2+qx+r}{ax+b}$
Differentiating with respect to $x$
$\Rightarrow f^{\prime }\left(x\right)=\frac{\left[\right(ax+b\left)\frac{d}{dx}\right(p{x}^2+qx+r)-(p{x}^2+qx+r\left)\frac{d}{dx}\right(ax+b\left)\right]}{(ax+b{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{(2px+q)(ax+b)-a(p{x}^2+qx+r)}{(ax+b{)}^2}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{2ap{x}^2+2bpx+qax+bq-ap{x}^2-qax-ar}{(ax+b{)}^2}$
$\therefore f^{\prime }\left(x\right)=\frac{ap{x}^2+2bpx+bq-ar}{(ax+b{)}^2}$