# How do you find the derivative of Y= ( 2x + 9 ) sqrt(x^2 - 4)?

Aug 4, 2015

$Y ' = \left(2 x + 9\right) \cdot \frac{x}{\sqrt{{x}^{2} - 4}} + 2 \cdot \sqrt{{x}^{2} - 4}$

#### Explanation:

Let $f \left(x\right) = 2 x + 9$ and $g \left(x\right) = \sqrt{{x}^{2} - 4}$. Their derivatives are:
$f ' \left(x\right) = 2$ and $g ' \left(x\right) = \frac{1}{2} {\left({x}^{2} - 4\right)}^{- \frac{1}{2}} \cdot 2 x = \frac{x}{\sqrt{{x}^{2} - 4}}$

The Product Rule says that

$Y ' = f \left(x\right) g ' \left(x\right) + f ' \left(x\right) g \left(x\right)$

$Y ' = \left(2 x + 9\right) \cdot \frac{x}{\sqrt{{x}^{2} - 4}} + 2 \cdot \sqrt{{x}^{2} - 4}$