# How do you differentiate g(x) = sin(3x)(x^2-4) using the product rule?

Jan 22, 2016

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

#### Explanation:

The product rule states that

$g ' \left(x\right) = \left({x}^{2} - 4\right) \frac{d}{\mathrm{dx}} \left[\sin \left(3 x\right)\right] + \sin \left(3 x\right) \frac{d}{\mathrm{dx}} \left[{x}^{2} + 4\right]$

Find each derivative.

The first will require the chain rule: $\frac{d}{\mathrm{dx}} \left[\sin \left(u\right)\right] = u ' \cdot \cos \left(u\right)$, and we have $u = 3 x$.

$\frac{d}{\mathrm{dx}} \left[\sin \left(3 x\right)\right] = \frac{d}{\mathrm{dx}} \left[3 x\right] \cdot \cos \left(3 x\right) = 3 \cos \left(3 x\right)$

The next simply requires power rule.

$\frac{d}{\mathrm{dx}} \left[{x}^{2} - 4\right] = 2 x$

Plug these back in to determine $g ' \left(x\right)$.

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