Explanation

(c)
$f\left( \frac{\pi}{4} \right) = \cos{\frac{\pi}{4\ }\cos{\frac{\pi}{2}\cos{\pi\cos{2\pi\cos{4\pi = 0}}}}}$ …(i)
Taking log on given function, we get
$\log{f(x) = \log{\cos{x + \log{\cos{2x + \log{\cos{4x + \log{\cos{8x + \log{\cos{16\ x}}}}}}}}}}}$
$\Rightarrow \ \frac{1}{f(x)}\ f^{'}(x) = \ - \tan{x - 2\tan{2x - 4\tan{4x - 8\tan{8x - 16\tan{16x}}}}}$
$\Rightarrow f^{'}(x) = \ - f(x)\lbrack\tan{x + 2\tan{2x + 4\tan{4x + 8\tan{8x + 16\tan{16x\rbrack}}}}}$
$\Rightarrow f^{'}\left( \frac{\pi}{4} \right)$
$= \ - f\left( \frac{\pi}{4} \right)\ \left\lbrack \tan{\frac{\pi}{4} + 2\tan{\frac{\pi}{2} + 4\tan{\pi + 8\tan{2\pi + 16\tan{4\pi}}}}} \right\rbrack$
$= \ 0\$ [using eq.(i)]