How do you differentiate g(x) = sqrt(x^2-x)cos3x using the product rule?

$g ' \left(x\right) = \frac{\left(\cos 3 x\right) \left(2 x - 1\right)}{\sqrt{{x}^{2} - x}} - 3 \sin \left(3 x\right) \sqrt{{x}^{2} - x}$
$g \left(x\right) = \left(\cos 3 x\right) \sqrt{{x}^{2} - x}$
$g ' \left(x\right) = \left(\cos 3 x\right) \times \frac{1}{2 \sqrt{{x}^{2} - x}} \times \left(2 x - 1\right) + \sqrt{{x}^{2} - x} \times \left(- 3\right) \left(\sin 3 x\right)$
$g ' \left(x\right) = \frac{\left(\cos 3 x\right) \left(2 x - 1\right)}{\sqrt{{x}^{2} - x}} - 3 \sin \left(3 x\right) \sqrt{{x}^{2} - x}$