Question

Differentiate the function with respect to x. 

cos x . cos 2x . cos 3x

Answer 1

Let, y = cos x · cos 2x · cos 3x           …(1)

Taking logarithm of both the sides,

log y = log (cos x · cos2x · cos 3x)

= log cos x + log cos 2 x + log cos 3x
    ....[∵ log m · n = log m + log n]

Differentiating both sides with respect to x,

${\displaystyle \frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}\log \cos x+\frac{d}{dx}\log \cos 2 x+\frac{d}{dx}\log \cos 3x}$

${\displaystyle \frac{1}{y} \frac{dy}{dx}=\frac{1}{\cos x}\frac{d}{dx}\cos x+\frac{1}{\cos 2 x}\frac{d}{dx}\cos 2\frac{x_1}{\cos 3 x}\frac{d}{dx}\cos 3 x}$

${\displaystyle =\frac{1}{\cos x}(-\sin x)+\frac{1}{\cos 2x}(-\sin 2x)\frac{d}{dx}\left(2x\right)+\frac{1}{\cos 3x}(-\sin 3x)\frac{d}{dx}\left(3x\right)}$

${\displaystyle =-\frac{\sin }{\cos x}-\frac{\sin 2x}{\cos 2x}\left(2\right)-\frac{\sin 3x}{\cos 3x}\left(3\right)}$

${\displaystyle =-\tan x-2 \tan 2x -3 \tan 3x =-(\tan x +2\tan 2x +3 \tan 3x)}$

${\displaystyle \therefore \frac{dy}{dx}=-y(\tan x +2\tan 2x +3 \tan 3x)}$

Putting the value of y from equation (1)

${\displaystyle \frac{dy}{dx}=-\cos x\cdot \cos 2 x\cdot \cos 3 x[\tan x+2 \tan 2 x +3 \tan 3 x]}$