How do you differentiate g(x) = sqrt(2x^2-1)cos3x using the product rule?

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

Explanation:

The formula for the product rule is

$\frac{d}{\mathrm{dx}} \left(u v\right) = u \frac{\mathrm{dv}}{\mathrm{dx}} + v \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

from the given $g \left(x\right) = \sqrt{2 {x}^{2} - 1} \cdot \cos 3 x$

We let $u = \sqrt{2 {x}^{2} - 1}$ and $v = \cos 3 x$

$\frac{d}{\mathrm{dx}} \left(u v\right) = u \frac{\mathrm{dv}}{\mathrm{dx}} + v \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$g ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(\sqrt{2 {x}^{2} - 1} \cdot \cos 3 x\right) = \left(\sqrt{2 {x}^{2} - 1}\right) \frac{d}{\mathrm{dx}} \left(\cos 3 x\right) + \left(\cos 3 x\right) \cdot \frac{d}{\mathrm{dx}} \sqrt{2 {x}^{2} - 1}$

$g ' \left(x\right) =$
$\left(\sqrt{2 {x}^{2} - 1}\right) \cdot \left(- 3 \sin 3 x\right) + \left(\cos 3 x\right) \cdot \frac{1}{2 \cdot \sqrt{2 {x}^{2} - 1}} \cdot \left(4 x - 0\right)$

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

God bless....I hope the explanation is useful.