Question

Differentiate the function given below w.r.t. $x:$

$\frac{1}{a{x}^2+bx+c}$

Answer 1

Let $y=\frac{1}{a{x}^2+bx+c}$

Differentiating with respect to $x,$

$\frac{dy}{dx}=\frac{(a{x}^2+bx+c)\frac{d\left(1\right)}{dx}-1\frac{d(a{x}^2+bx+c)}{dx}}{(a{x}^2+bx+c{)}^2}$

$[\text{quotient rule}\frac{d\left(\frac{u}{v}\right)}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}]$

$\Rightarrow \frac{dy}{dx}=\frac{0.(a{x}^2+bx+c)-1.(2ax+b)}{(a{x}^2+bx+c{)}^2}$

$\Rightarrow \frac{dy}{dx}=\frac{-(2ax+b)}{(a{x}^2+bx+c{)}^2}$