# How do you differentiate f(x)= (x^2-5x+2)/ (6x+1 ) using the quotient rule?

May 3, 2018

$f ' \left(x\right) = \frac{6 {x}^{2} + 2 x - 17}{6 x + 1} ^ 2$

#### Explanation:

I've always remembered this rule by the saying "Low d-high minus high d-low, all over the square of what's below." This translates into the following procedure:

Suppose you have some function $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$, where $g \left(x\right)$ and $h \left(x\right)$ are some functions of $x$. Then the derivative of $f \left(x\right)$ is $f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{{h}^{2} \left(x\right)}$.

Solving the problem you presented, we have:

$f ' \left(x\right) = \frac{\left(6 x + 1\right) \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} - 5 x + 2\right) - \left({x}^{2} - 5 x + 2\right) \cdot \frac{d}{\mathrm{dx}} \left(6 x + 1\right)}{6 x + 1} ^ 2$
$= \frac{\left(6 x + 1\right) \left(2 x - 5\right) - \left({x}^{2} - 5 x + 2\right) \left(6\right)}{6 x + 1} ^ 2$
$= \frac{\left(12 {x}^{2} + 2 x - 30 x - 5\right) - \left(6 {x}^{2} - 30 x + 12\right)}{6 x + 1} ^ 2$
$= \frac{12 {x}^{2} - 28 x - 5 - 6 {x}^{2} + 30 x - 12}{6 x + 1} ^ 2$
$= \frac{6 {x}^{2} + 2 x - 17}{6 x + 1} ^ 2 = f ' \left(x\right)$