# How do you differentiate f(x) = (5x)/(3x^2-4x+6) using the quotient rule?

Jan 20, 2016

$f ' \left(x\right) = \frac{- 15 {x}^{2} + 30}{3 {x}^{2} - 4 x + 6} ^ 2$

#### Explanation:

For a function $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$, the quotient rule states that the derivative of the function is

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

In this scenario, we know the following:

$g \left(x\right) = 5 x \textcolor{w h i t e}{\times \times} \implies \textcolor{w h i t e}{\times \times} g ' \left(x\right) = 5$

$h \left(x\right) = 3 {x}^{2} - 4 x + 6 \textcolor{w h i t e}{\times \times} \implies \textcolor{w h i t e}{\times \times} h ' \left(x\right) = 6 x - 4$

Thus, plugging these into the quotient rule formula,

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

Simplify.

$f ' \left(x\right) = \frac{15 {x}^{2} - 20 x + 30 - 30 {x}^{2} + 20 x}{3 {x}^{2} - 4 x + 6} ^ 2$

$f ' \left(x\right) = \frac{- 15 {x}^{2} + 30}{3 {x}^{2} - 4 x + 6} ^ 2$