# How do you differentiate f(x)= (3x^2-4 )/ (2x- 1 ) using the quotient rule?

Nov 4, 2016

Derivative of $f \left(x\right)$ comes to be $\frac{6 x \left(2 x - 1\right) - 2 \left(3 {x}^{2} - 4\right)}{2 x - 1} ^ 2$.

#### Explanation:

Let, $y = f \left(x\right) = \frac{3 {x}^{2} - 4}{2 x - 1} .$
Therefore, Differentiating $y$ w.r.t $x$ comes to be,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{v \cdot \frac{\mathrm{du}}{\mathrm{dx}} - u \cdot \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(2 x - 1\right) \cdot \frac{d}{\mathrm{dx}} \left(3 {x}^{2} - 4\right) - \left(3 {x}^{2} - 4\right) \cdot \frac{d}{\mathrm{dx}} \left(2 x - 1\right)}{2 x - 1} ^ 2$

:.dy/dx=((2x-1)*(3*2*x^(2-1)-0)-(3x^2-4)*(2-0)
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 x \left(2 x - 1\right) - 2 \left(3 {x}^{2} - 4\right)}{2 x - 1} ^ 2$. (answer).