Question

If $f\left(x\right)=\frac{ax+b}{p{x}^2+qx+r^{\prime }}$ then $f^{\prime }\left(x\right)$ is

Answer 1

$f\left(x\right)=\frac{ax+b}{p{x}^2+qx+r}{f}^{{}^{\prime }}\left(x\right)=\frac{a\left(p{x}^2+qx+r\right)-(2px+q)(ax+b)}{{\left(p{x}^2+qx+r\right)}^2}{f}^{{}^{\prime }}\left(x\right)=\frac{ap{x}^2+aqx+ar-2ap{x}^2-2pbx-qax-bq}{{\left(p{x}^2+qx+r\right)}^2}{f}^{{}^{\prime }}\left(x\right)=\frac{-ap{x}^2+ar-2pbx-bq}{{\left(p{x}^2+qx+r\right)}^2}$