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

Apr 25, 2018

$g ' \left(x\right) = 18 \left(2 x + 3\right) \left(\cos 6 x\right) + 6 \left(\sin 6 x\right)$
$g \left(x\right) = \left(6 x + 9\right) \left(\sin 6 x\right)$
$g ' \left(x\right) = \left(6 x + 9\right) \times 6 \left(\cos 6 x\right) + \left(\sin 6 x\right) \times 6$
$g ' \left(x\right) = 18 \left(2 x + 3\right) \left(\cos 6 x\right) + 6 \left(\sin 6 x\right)$