How do you differentiate g(x) = sin(6x)(x^2-4) using the product rule?

Feb 23, 2016

$g ' \left(x\right) = 2 x \sin 6 x + 6 \left({x}^{2} - 4\right) \cos 6 x$

Explanation:

using the $\textcolor{b l u e}{\text{ Product rule and chain rule }}$

If g(x) = f(x)h(x) then g'(x) = f(x)h'(x) + h(x)f'(x)

$\textcolor{red}{\text{ chain rule }} : \frac{d}{\mathrm{dx}} \left[f \left(g \left(x\right)\right)\right] = f ' \left(g \left(x\right)\right) g ' \left(x\right)$

hence g'(x) $= \sin 6 x \frac{d}{\mathrm{dx}} \left({x}^{2} - 4\right) + \left({x}^{2} - 4\right) \frac{d}{\mathrm{dx}} \left(\sin 6 x\right)$

$= \sin 6 x \left(2 x\right) + \left({x}^{2} - 4\right) \left(\cos 6 x\right) \frac{d}{\mathrm{dx}} \left(6 x\right)$

$= 2 x \sin 6 x + \left({x}^{2} - 4\right) \cos 6 x \left(6\right)$

$= 2 x \sin 6 x + 6 \left({x}^{2} - 4\right) \cos 6 x$