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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} $

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  • $\log mn=\log m+\log n$
  • $\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]$
$\quad=-\cos x.\cos 2x.\cos 3x[\tan x+2\tan 2x+3\tan 3x]$
answered May 8, 2013 by sreemathi.v

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