Solve this

Let $y=e^{3 x} \cos 2 x, z=e^{3 x}$ and $w=\cos 2 x$

Formula :

$\frac{\mathrm{d}\left(\mathrm{e}^{\mathrm{x}}\right)}{\mathrm{dx}}=\mathrm{e}^{\mathrm{x}}$ and $\frac{\mathrm{d}(\cos \mathrm{x})}{\mathrm{dx}}=-\sin \mathrm{x}$

According to the product rule of differentiation

$d y / d x=w \times \frac{d z}{d x}+z \times \frac{d w}{d x}$

$=\left[\cos 2 x \times\left(3 \times e^{3 x}\right)\right]+\left[e^{3 x} \times(-2 \sin 2 x)\right]$

$=e^{3 x} \times[3 \cos 2 x-2 \sin 2 x]$