# How do you differentiate g(x) = (6x+9)sin(3x) using the product rule?

Oct 20, 2017

$g ' \left(x\right) = 3 \left(6 x + 9\right) \cos \left(3 x\right) + 6 \sin \left(3 x\right)$

#### Explanation:

First, we recall product rule :

$f \left(x\right) = a \cdot b$
$f ' \left(x\right) = a \cdot \frac{\mathrm{db}}{\mathrm{dx}} + b \cdot \frac{\mathrm{da}}{\mathrm{dx}}$

From the above question, we can think of $\left(6 x + 9\right)$as $a$ and $\sin \left(3 x\right)$ as $b$. So, we have the following:

$g ' \left(x\right) = \left(6 x + 9\right) \cdot d \frac{\sin \left(3 x\right)}{\mathrm{dx}} + \sin \left(3 x\right) \cdot d \frac{6 x + 9}{\mathrm{dx}}$
$= \left(6 x + 9\right) \cdot \cos \left(3 x\right) d \frac{3 x}{\mathrm{dx}} + \sin \left(3 x\right) \cdot 6$
$= 3 \left(6 x + 9\right) \cos \left(3 x\right) + 6 \sin \left(3 x\right)$