Explanation

(b)
We know that, $\cos{A\cos{2A{\cos\ 2^{2}}{A\ldots{cos2}^{n - 1}{A =}}}}\frac{sin(2^{n}A)}{2^{n}\sin A}$
$\therefore\cos{x\cos{2x\cos{4x\cos{8x\cos{16x = \frac{\sin{32\ x}}{32\sin x}}}}}}$
$\Rightarrow f(x) = \frac{1}{32}.\frac{\sin{32x}}{\sin x}$
$\therefore f^{'}(x) = \frac{1}{32} \times \frac{\sin{x(32\cos{32x) - \sin{32x\ \cos\ x}}}}{\sin^{2}x}$
$\Rightarrow f^{'}\left( \frac{\pi}{4} \right) = \frac{1}{\sin\frac{\pi}{4}} = \sqrt{2}$