Question

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

Answer 1

Simplification of given data

Let $y=\cos x.\cos 2x.\cos 3x$

Taking $\log$ on both sides, we get

$\log y=\log (\cos x.\cos 2x.\cos 3x)$

$\log y=\log \cos x+\log \cos 2x+\log \cos 3x$

$(\log ab=\log a+\log b)$

Differentiating both sides w.r.t.$x$, we get,

$\frac{d\left(\log y\right)}{dx}=\frac{d\left(\log \right(\cos x)+\log (\cos 2x)+\log (\cos 3x\left)\right)}{dx}$

Multiplying and dividing by $dy$

$\frac{d\left(\log y\right)}{dy}.\frac{dy}{dx}=\frac{d\left(\log \right(\cos x\left)\right)}{dx}+\frac{d\left(\log \right(\cos 2x\left)\right)}{dx}+\frac{d\left(\log \right(\cos 3x\left)\right)}{dx}$

$(\text{using chain rule}\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx})$

$\frac{d\left(\log y\right)}{dy}.\frac{dy}{dx}=\frac{1}{\cos x}.\frac{d\left(\cos x\right)}{dx}+\frac{1}{\cos 2x}.\frac{d\left(\cos 2x\right)}{dx}+\frac{1}{\cos 3x}.\frac{d\left(\cos 3x\right)}{dx}$

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

$\frac{1}{y}.\frac{dy}{dx}=\frac{-\sin x}{\cos x}-\frac{\sin 2x}{\cos 2x}.2-\frac{\sin 3x}{\cos 3x}.3$

$\frac{1}{y}.\frac{dy}{dx}=-(\tan x+2\tan 2x+3\tan 3x)$

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

Substituing the value of

$y=\cos x.\cos 2x.\cos 3x$, we get

$\frac{dy}{dx}=-\cos x.\cos 2x.\cos 3x(\tan x+2\tan 2x+3\tan 3x)$