# How do you differentiate g(x) = sqrt(x^3-4)cos2x using the product rule?

Jan 12, 2016

$g ' \left(x\right) = \frac{3 {x}^{2} \cdot \cos 2 x}{2 \sqrt{{x}^{3} - 4}} - 2 \sin \left(2 x\right) \cdot \sqrt{{x}^{3} - 4}$

#### Explanation:

For $g \left(x\right) = u \left(x\right) v \left(x\right)$, the product rule states that the derivative can be computed as follows:

$g ' \left(x\right) = u ' \left(x\right) v \left(x\right) + u \left(x\right) v ' \left(x\right)$

$u \left(x\right) = \sqrt{{x}^{3} - 4} \text{ }$ and

$v \left(x\right) = \cos 2 x$

So, first thing to do is compute the derivatives of $u \left(x\right)$ and $v \left(x\right)$ using the chain rule:

$u \left(x\right) = \sqrt{{x}^{3} - 4} = {\left({x}^{3} - 4\right)}^{\frac{1}{2}}$
$\textcolor{w h i t e}{\times \times \times \times \times x} \implies u ' \left(x\right) = \frac{1}{2} {\left({x}^{3} - 4\right)}^{- \frac{1}{2}} \cdot 3 {x}^{2} = \frac{1}{2 \sqrt{{x}^{3} - 4}} \cdot 3 {x}^{2}$

$v \left(x\right) = \cos 2 x \textcolor{w h i t e}{\times} \implies v ' \left(x\right) = - \sin \left(2 x\right) \cdot 2$

Now, the only thing left to do is apply the formula!

$g ' \left(x\right) = u ' \left(x\right) \cdot v \left(x\right) + u \left(x\right) \cdot v ' \left(x\right)$

$\textcolor{w h i t e}{\times \times} = \frac{1}{2 \sqrt{{x}^{3} - 4}} \cdot 3 {x}^{2} \cdot \cos 2 x - 2 \sin \left(2 x\right) \cdot \sqrt{{x}^{3} - 4}$

$\textcolor{w h i t e}{\times \times} = \frac{3 {x}^{2} \cos 2 x}{2 \sqrt{{x}^{3} - 4}} - 2 \sin \left(2 x\right) \cdot \sqrt{{x}^{3} - 4}$