Find the derivative of $\dfrac{2x + 1}{3x - 4}$

The correct answer is $\dfrac{- 11}{(3x - 4)^2}$

As per Quotient Rule,

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

$ ∴ \dfrac{dy}{dx}$ = $\dfrac{(3x - 4) D(2x + 1) - (2x + 1) D(3x - 4)}{(3x - 4)^2}$

= $\dfrac{(3x - 4) (2) - (2x + 1)(3)}{(3x - 4)^2}$

= $ \dfrac{6x - 8 - 6x -3}{(3x - 4)^2}$

= $\dfrac{- 11}{(3x - 4)^2}$