Find $\frac{dy}{dx}$ where $y=\cos x\cos 2x\cos 3x\cos 4x\cos 5x$.

$\cos x\cos 2x\cos 3x\cos 4x\cos 5x(-\tan x-2\tan 2x-3\tan 3x-4\tan 4x-5\tan 5x)$ $(-\tan x-2\tan 2x-3\tan 3x-4\tan 4x-5\tan 5x)$ $\cos x\cos 2x\cos 3x\cos 4x\cos 5x$ $\log(\cos x)+\log(\cos 2x)+\log(\cos 3x)+\log(\cos 4x)+\log(\cos 5x)$

$\cos x\cos 2x\cos 3x\cos 4x\cos 5x(-\tan x-2\tan 2x-3\tan 3x-4\tan 4x-5\tan 5x)$

$y=\cos x\cos 2x\cos 3x\cos 4x\cos 5x$. Taking log in both sides we get $\log y=\log cos x+\log cos 2x+\log cos 3x+\log cos 4x+\log cos 5x$. Now differentiating both sides w.r.to x we get $1/y\frac{dy}{dx}=\frac{d}{dx}(\log cos x)+\frac{d}{dx}(\log cos 2x)+\frac{d}{dx}(\log cos 3x)+\frac{d}{dx}(\log cos 4x)+\frac{d}{dx}(\log cos 5x)=\frac{-\sin x}{cos x}+\frac{-2\sin 2x}{\cos 2x}+\frac{-3\sin 3x}{\cos 3x}+\frac{-4\sin 4x}{\cos 4x}+\frac{-5\sin 5x}{\cos 5x}=-\tan x-2\tan 2x-3\tan 3x-4\tan 4x-5\tan 5x$. Hence $\frac{dy}{dx}=y(-\tan x-2\tan 2x-3\tan 3x-4\tan 4x-5\tan 5x)=\cos x\cos 2x\cos 3x\cos 4x\cos 5x(-\tan x-2\tan 2x-3\tan 3x-4\tan 4x-5\tan 5x)$