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

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

As per Quotient Rule,


$\dfrac{dy}{dx}$ = $\dfrac{v \dfrac{du}{dx} - u \dfrac{dv}{dx}}{v^2}$


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


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


= $\dfrac{4x^2 + 2x - 2x^2 + 6 )}{(2x + 1)^2}$


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