How do you differentiate f(x)=e^(3x)cos2x using the product rule?

$f ' \left(x\right) = {e}^{3 x} \cdot \left[3 \cdot \cos 2 x - 2 \cdot \sin 2 x\right]$

Explanation:

Given $f \left(x\right) = {e}^{3 x} \cdot \cos \left(2 x\right)$

Use the formula
$\frac{d}{\mathrm{dx}} \left(u v\right) = u \cdot \frac{d}{\mathrm{dx}} \left(v\right) + v \cdot \frac{d}{\mathrm{dx}} \left(u\right)$
Let $u = {e}^{3 x}$ and $v = \cos \left(2 x\right)$

$\frac{d}{\mathrm{dx}} \left(u v\right) = u \cdot \frac{d}{\mathrm{dx}} \left(v\right) + v \cdot \frac{d}{\mathrm{dx}} \left(u\right)$
$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left({e}^{3 x} \cdot \cos \left(2 x\right)\right) = {e}^{3 x} \cdot \frac{d}{\mathrm{dx}} \left(\cos \left(2 x\right)\right) + \cos \left(2 x\right) \cdot \frac{d}{\mathrm{dx}} \left({e}^{3 x}\right)$

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

$f ' \left(x\right) = {e}^{3 x} \cdot \left(- \sin \left(2 x\right)\right) \left(2\right) + \cos \left(2 x\right) {e}^{3 x} \left(3\right)$

simplify by factoring common factors

$f ' \left(x\right) = {e}^{3 x} \cdot \left[3 \cdot \cos 2 x - 2 \cdot \sin 2 x\right]$

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