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# Differentiate the functions w.r.t. x. cos x cos 2x cos 3x

Q1   Differentiate the functions w.r.t. x.  $\cos x . \cos 2x .\cos 3x$

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Given function is
$y=\cos x . \cos 2x .\cos 3x$
Now, take log on both sides
$\log y=\log (\cos x . \cos 2x .\cos 3x)\\ \log y = \log \cos x + \log \cos 2x + \log \cos 3x$
Now, differentiation w.r.t. x
$\log y=\log (\cos x . \cos 2x .\cos 3x)\\ \frac{d(\log y )}{dx} = \frac{\log \cos x}{dx} + \frac{\log \cos 2x}{dx} + \frac{\log \cos 3x}{dx}\\ \frac{1}{y}.\frac{dy}{dx} = (-\sin x)\frac{1}{\cos x}+(-2\sin 2x)\frac{1}{\cos 2x}+(-3\sin3x).\frac{1}{\cos3x}\\ \frac{1}{y}\frac{dy}{dx} = -(\tan x+\tan 2x+\tan 3x) \ \ \ \ \ \ (\because \frac{\sin x }{\cos x} =\tan x)\\ \frac{dy}{dx}=-y(\tan x+\tan 2x+\tan 3x)\\ \frac{dy}{dx}= -\cos x\cos 2x\cos 3x(\tan x+\tan 2x+\tan 3x)$
There, the answer is  $-\cos x\cos 2x\cos 3x(\tan x+\tan 2x+\tan 3x)$

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