If $f(x)=\cos x \cos 2 x \cos 4 x \cos 8 x \cos 16 x$, then $f'\left(\frac{\pi}{4}\right)$ is
Answer & explanation
Correct answer: option 1
We have,
$f(x)=\cos x \cos 2 x \cos 4 x \cos 8 x \cos 16 x$
$f\left(\frac{\pi}{4}\right)=0$
$\log f(x)=\log \cos x +\log \cos 2 x +\log \cos 4 x +\log \cos 8 x +\log \cos 16 x$
differentiating wrt x
$\frac{df(x)}{dx}=-f(x)(\tan x+2\tan 2x+4\tan 4x+8\tan 8x+16\tan 16x)$
so $f'\left(\frac{\pi}{4}\right)=0$ as $f\left(\frac{\pi}{4}\right)=0$