# Differentiate the functions given in w.r.t. $x :$ $\cos \: x \: . \cos \: 2x \:. \cos \: 3x$

$\begin{array}{1 1} -\cos x.\sin 3x.\cos 3x[\tan x+2\tan 2x+3\tan 3x] \\ -\cos x.\cos 2x.\cos 3x[\tan x+2\tan 2x+3\tan 3x] \\ -\cos x.\cos 3x.\cos 3x[\tan x+2\tan 2x+3\tan 3x] \\ -\cos x.\cos 3x.\cos 3x[\tan x+2\cot 2x+3\tan 3x]\end{array}$

Toolbox:
• $\log mn=\log m+\log n$
• $\large\frac{d}{dx}$$(\log x)=\large\frac{1}{x} Step 1: Let y=\cos x.\cos 2x.\cos 3x Taking \log of both sides \log y=\log(\cos x.\cos 2x.\cos 3x) We know that \log mn=\log m+\log n \log y=\log\cos x+\log\cos 2x+\log\cos 3x Step 2: Differentiating both sides with respect to x \large\frac{1}{y}\frac{dy}{dx}=\large\frac{1}{\cos x}\frac{d}{dx}$$(\cos x)+\large\frac{1}{\cos 2x}\frac{d}{dx}$$(\cos 2x)+\large\frac{1}{\cos 3x}\frac{d}{dx}$$(\cos 3x)$
$\qquad=\large\frac{1}{\cos x}$$(-\sin x)+\large\frac{1}{\cos 2x}$$(\sin 2x.2)+\large\frac{1}{\cos 3x}$$(\sin 3x.3) \Rightarrow \large\frac{-\sin x}{\cos x}+\large\frac{-2\sin 2x}{\cos 2x}+\large\frac{-3\sin 3x}{\cos 3x} \Rightarrow -[\tan x+2\tan 2x+3\tan 3x] Step 3: \large\frac{1}{y}\frac{dy}{dx}$$=-[\tan x+2\tan 2x+3\tan 3x]$
$\large\frac{dy}{dx}$$=-y.[\tan x+2\tan 2x+3\tan 3x]$
$\quad=-\cos x.\cos 2x.\cos 3x[\tan x+2\tan 2x+3\tan 3x]$