Question

If f(x) $=\frac{1}{a{x}^2+bx+c}$ then f '(x) $=$

• A. $\frac{-(2ax+b)}{(a{x}^2+bx+c{)}^2}$
• B. $2ax+b$
• C. $\frac{2ax+b}{a{x}^2+bx+c}$
• D. None

Answer 1

Answer: (A)

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

Let $f\left(x\right)=\frac{1}{a{x}^2+bx+c}$
$\Rightarrow f^{\prime }\left(x\right)=\frac{(a{x}^2+bx+c)\frac{d}{dx}1-1.\frac{d}{dx}(a{x}^2+bx+c)}{(a{x}^2+bx+c^2)}$
$\therefore f^{\prime }\left(x\right)=\frac{(a{x}^2+bx+c).0-(2x+b)}{(a{x}^2+bx+c{)}^2}=\frac{(-2ax+b)}{(a{x}^2+bx+c{)}^2}$