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

Sep 23, 2014

$y ' = \frac{g \left(x\right) f ' \left(x\right) - f \left(x\right) g ' \left(x\right)}{g \left(x\right)} ^ 2$

$y = f \frac{x}{g} \left(x\right) = \frac{2 {x}^{4} - 3 x}{4 x - 1}$

$f ' \left(x\right) = 8 {x}^{3} - 3$

$g ' \left(x\right) = 4$

${\left(g \left(x\right)\right)}^{2} = {\left(4 x - 1\right)}^{2}$

$y ' = \frac{\left(4 x - 1\right) \left(8 {x}^{3} - 3\right) - \left(2 {x}^{4} - 3 x\right) \left(4\right)}{4 x - 1} ^ 2$

$y ' = \frac{32 {x}^{4} - 12 x - 8 {x}^{3} + 3 - 8 {x}^{4} + 12 x}{4 x - 1} ^ 2$

Simplify for combining like terms.

$S o l u t i o n \to y ' = \frac{24 {x}^{4} - 8 {x}^{3} + 3}{4 x - 1} ^ 2$