Question

Find the derivative of the following function:
f(x)= $\frac{1}{a{x}^2+bx+c}$

Answer 1

Here $f\left(x\right)=\frac{1}{a{x}^2+bx+c}$

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

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

$=\frac{(a{x}^2+bx+c)\left(0\right)-1(2ax+b)}{(a{x}^2+bx+c{)}^2}$

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