Question

If $f(x) = \cos x.\cos 2x.\cos 4x.\cos 8x.\cos 16x$ then

$f^{'}\left( \frac{\pi}{4} \right)$ is

• A. $\sqrt{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. 0
• D. None of these

Answer 1

Answer: (A)

$\sqrt{2}$

Answer 2

$f(x) = \frac{2\sin x.\cos x.\cos 2x.\cos 4x.\cos 8x.\cos 16x}{2\sin x} = \frac{\sin 32x}{2^{5}\sin x}$

$\therefore$ $f^{'}(x) = \frac{1}{32}.\frac{32\cos 32x.\sin x - \cos x.\sin 32x}{\sin^{2}x}$

$\therefore$ $f^{'}\left( \frac{\pi}{4} \right) = \frac{32.\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}.0}{32.\left( \frac{1}{\sqrt{2}} \right)^{2}} = \sqrt{2}$.