$cos{\rm{ }}x \cdot cos{\rm{ }}2x \cdot cos{\rm{ }}3x$

Let $y = cos{\rm{ }}x \cdot cos{\rm{ }}2x \cdot cos{\rm{ }}3x$

By taking log on both sides , we get

$\log y = \log (\cos {\rm{ }}x \cdot cos{\rm{ }}2x \cdot cos{\rm{ }}3x)$

Then,
$log\,y\;log\left( {cosx} \right) + log\left( {cos2x} \right) + log\left( {cos3x} \right)$ ...(i)

On differentiating (i) both sides w.r.t. x, we get
$\cfrac{1}{y}.\cfrac{{dy}}{{dx}}$

$= \cfrac{1}{{\cos x}}( - \sin x) + \cfrac{1}{{\cos 2x}}( - \sin 2x)(2) + \cfrac{1}{{\cos 3x}}( - \sin 3x)(3)$

$\Rightarrow$ $\cfrac{1}{y}\cfrac{{dy}}{{dx}} = - \tan x - 2\tan 2x - 3\tan 3x$

therefore, $\cfrac{{dy}}{{dx}} = - \cos x \cdot \cos 2x \cdot \cos 3x(\tan x + 2\tan 2x + 3\tan 3x)$