Find the derivative of $\dfrac{e^{-3x}}{x^2 + 1}$

The correct answer is $\dfrac{- (3x^2 + 2x + 3)e^{-3x}}{(x^2 + 1)^2}$

As per Quotient Rule,
$\dfrac{dy}{dx}$ = $\dfrac{v \dfrac{du}{dx} - u \dfrac{dv}{dx}}{v^2}$

$ ∴ \dfrac{dy}{dx}$ = $\dfrac{(x^2 + 1) D(e^{-3x}) - (e^{-3x}) D(x)}{(x^2 + 1)^2}$

= $\dfrac{(x^2 + 1) (-3 e^{-3x}) - (e^{-3x})(2x)}{(x^2 + 1)^2}$

= $\dfrac{- (3x^2 + 3 + 2x)e^{-3x}}{(x^2 + 1)^2}$

= $\dfrac{- (3x^2 + 2x + 3)e^{-3x}}{(x^2 + 1)^2}$