# How do you differentiate (x^2 -6x + 9 )/ sqrt(x-3) using the quotient rule?

Dec 27, 2015

$f ' \left(x\right) = \frac{\left(2 x - 6\right) \sqrt{x - 3} - \left({x}^{2} - 6 x + 9\right) \left(\frac{1}{2 \sqrt{x - 3}}\right)}{x - 3}$

#### Explanation:

Let $f \left(x\right) = \frac{{x}^{2} - 6 x + 9}{\sqrt{x - 3}}$.

The quotient rule tells us that the derivative of $\frac{u \left(x\right)}{v \left(x\right)}$ is $\frac{u ' \left(x\right) v \left(x\right) - u \left(x\right) v ' \left(x\right)}{v {\left(x\right)}^{2}}$. Here, let $u \left(x\right) = {x}^{2} - 6 x + 9$ and $v \left(x\right) = \sqrt{x - 3}$. So $u ' \left(x\right) = 2 x - 6$ and $v ' \left(x\right) = \frac{1}{2 \sqrt{x - 3}}$.

We now apply the quotient rule.

$f ' \left(x\right) = \frac{\left(2 x - 6\right) \sqrt{x - 3} - \left({x}^{2} - 6 x + 9\right) \left(\frac{1}{2 \sqrt{x - 3}}\right)}{x - 3}$