# How do you differentiate g(x) = (5x^6 - 4) sin(6x) using the product rule?

Oct 8, 2016

The product rule is $\frac{d}{\mathrm{dx}} \left(u v\right) = u \frac{\mathrm{dv}}{\mathrm{dx}} + \frac{\mathrm{du}}{\mathrm{dx}} v$

Let $u \left(x\right) = 5 {x}^{6} - 4 \mathmr{and} v \left(x\right) = \sin 6 x$

Then, $\frac{\mathrm{du}}{\mathrm{dx}} = \left(5 {x}^{5}\right) 6 = 30 {x}^{5}$
And, $\frac{\mathrm{dv}}{\mathrm{dx}} = \left(\cos 6 x\right) 6 = 6 \cos 6 x$

So the product rue gives:

$\frac{\mathrm{dg}}{\mathrm{dx}} = \left(5 {x}^{6} - 4\right) \left(6 \cos 6 x\right) + \left(30 {x}^{5}\right) \left(\sin 6 x\right)$
$\therefore \frac{\mathrm{dg}}{\mathrm{dx}} = 30 {x}^{6} \cos 6 x - 24 \cos 6 x + 30 {x}^{5} \sin 6 x$