How do you differentiate (3x^2 -5x )/ (2x-1) using the quotient rule?

Jun 2, 2018

Answer:

$y ' = \frac{6 {x}^{2} - 6 x + 5}{2 x - 1} ^ 2$

Explanation:

Using the Quotient rule we get
$y ' = \frac{\left(6 x - 5\right) \left(2 x - 1\right) - \left(3 {x}^{2} - 5 x\right) \cdot 2}{2 x - 1} ^ 2$
and this is

$y ' = \frac{12 {x}^{2} - 10 x - 6 x + 5 - 6 {x}^{2} + 10 x}{2 x - 1} ^ 2$
this is

$y ' = \frac{6 {x}^{2} - 6 x + 5}{2 x - 1} ^ 2$