# How do you differentiate f(x)=(2x^3)(ln3x)(sinx)  using the product rule?

Aug 15, 2017

$f ' \left(x\right) = 6 {x}^{2} \setminus \ln 3 x \setminus \sin x + 2 {x}^{2} \setminus \sin x + 2 {x}^{3} \setminus \ln 3 x \setminus \cos x$

#### Explanation:

We have:

$f \left(x\right) = \left(2 {x}^{3}\right) \left(\ln 3 x\right) \left(\sin x\right)$

We need to use the triple product rule, giving

$f ' \left(x\right) = \left(\frac{d}{\mathrm{dx}} 2 {x}^{3}\right) \left(\ln 3 x\right) \left(\sin x\right) + \left(2 {x}^{3}\right) \left(\frac{d}{\mathrm{dx}} \ln 3 x\right) \left(\sin x\right) + \left(2 {x}^{3}\right) \left(\ln 3 x\right) \left(\frac{d}{\mathrm{dx}} \sin x\right)$

$\text{ } = \left(6 {x}^{2}\right) \left(\ln 3 x\right) \left(\sin x\right) + \left(2 {x}^{3}\right) \left(\frac{1}{3 x} \frac{d}{\mathrm{dx}} 3 x\right) \left(\sin x\right) + \left(2 {x}^{3}\right) \left(\ln 3 x\right) \left(\cos x\right)$

$\text{ } = \left(6 {x}^{2}\right) \left(\ln 3 x\right) \left(\sin x\right) + \left(2 {x}^{3}\right) \left(\frac{3}{3 x}\right) \left(\sin x\right) + \left(2 {x}^{3}\right) \left(\ln 3 x\right) \left(\cos x\right)$

$\text{ } = 6 {x}^{2} \setminus \ln 3 x \setminus \sin x + 2 {x}^{2} \setminus \sin x + 2 {x}^{3} \setminus \ln 3 x \setminus \cos x$