# How do you differentiate y=(4x^2-3x+1)/(5x-3)?

Feb 27, 2017

Use the quotient rule: $\left(\frac{u}{v}\right) ' = \frac{v u ' - u v '}{v} ^ 2$

$f \left(x\right) ' = \frac{4 \left(5 {x}^{2} - 6 x + 1\right)}{5 x - 3} ^ 2$

#### Explanation:

Use the quotient rule: $\left(\frac{u}{v}\right) ' = \frac{v u ' - u v '}{v} ^ 2$

Let $u = 4 {x}^{2} - 3 x + 1$, so $u ' = 8 x - 3$

Let $v = 5 x - 3$, so $v ' = 5$

$f \left(x\right) ' = \frac{\left(5 x - 3\right) \left(8 x - 3\right) - \left(4 {x}^{2} - 3 x + 1\right) \left(5\right)}{5 x - 3} ^ 2$

Distribute the numerator:

$f \left(x\right) ' = \frac{40 {x}^{2} - 15 x - 24 x + 9 - 20 {x}^{2} + 15 x - 5}{5 x - 3} ^ 2$

Simplify numerator:

$f \left(x\right) ' = \frac{20 {x}^{2} - 24 x + 4}{5 x - 3} ^ 2 = \frac{4 \left(5 {x}^{2} - 6 x + 1\right)}{5 x - 3} ^ 2$