Question

Differentiate the function w.r.t. $x$.
$\cos x.\cos 2x.\cos 3x$

Answer 1

Let $y=\cos x.\cos 2x.\cos 3x$
Taking logarithm on both the sides, we obtain
$\log y=\log (\cos x.\cos 2x.\cos 3x)=\log \left(\cos x\right)+\log \left(\cos 2x\right)+\log \left(\cos 3x\right)$
Differentiating both sides with respect to $x$, we obtain
$\frac{1}{y}\frac{dy}{dx}=\frac{1}{\cos x}.\frac{d}{dx}\left(\cos x\right)+\frac{1}{\cos 2x}.\frac{d}{dx}\left(\cos 2x\right)+\frac{1}{\cos 3x}.\frac{d}{dx}\left(\cos 3x\right)$
$\Rightarrow \frac{dy}{dx}=y[-\frac{\sin x}{\cos x}-\frac{\sin 2x}{\cos 2x}.\frac{d}{dx}(2x)-\frac{\sin 3x}{\cos 3x}.\frac{d}{dx}(3x\left)\right]$
$\Rightarrow \frac{dy}{dx}=-\cos x.\cos 2x.\cos 3x[\tan x+2\tan 2x+3\tan 3x]$