Solution:

Let’s take $y=\cos x \cos 2 x \cos 3 x$

On both the sides taking log, we get

$\log y=\log (\cos x \cos 2 x \cos 3 x)$

$=\log \cos x+\log \cos 2 x+\log \cos 3 x$

So now,

$\frac{d}{d x} \log y=\frac{d}{d x} \log \cos x+\frac{d}{d x} \log \cos 2 x+\frac{d}{d x} \log \cos 3 x$

$\frac{1}{y} \cdot \frac{d y}{d x}=\frac{1}{\cos x} \frac{d}{d x} \cos x+\frac{1}{\cos 2 x} \frac{d}{d x} \cos 2 x+\frac{1}{\cos 3 x} \frac{d}{d x} \cos 3 x$

$\frac{1}{y} \cdot \frac{d y}{d x}=\frac{1}{\cos x}(-\sin x)+\frac{1}{\cos 2 x}(-\sin 2 x) \frac{d}{d x} 2 x+\frac{1}{\cos 3 x}(-\sin 3 x) \frac{d}{d x} 3 x$

$\frac{1}{y} \cdot \frac{d y}{d x}=-\tan x-(\tan 2 x) 2-\tan 3 x(3)$

$\frac{d y}{d x}=-y(\tan x+2 \tan 2 x+3 \tan 3 x)$

$\frac{d y}{d x}=-\cos x \cos 2 x \cos 3 x(\tan x+2 \tan 2 x+3 \tan 3 x) \text { [using value of } y$