Question

If $f^{(x)}=\cos x \cos 2 x \cos 4 x \cos 8 x \cos 16 x$, then $f^{\prime}\left(\frac{\pi}{4}\right)$ is (where $\left.f^{\prime}(x)=\frac{ d }{d x } f^{(x)}\right)$

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

Answer 1

Answer: (A)

We have, $f\left(x\right)=cosxcos2xcos4xcos8xcos16x$ $\Rightarrow f\left(x\right)=\frac{2 sin x cos x cos 2 x cos 4 x cos 8 x cos 16 x}{2 sin x}$
$=\frac{sin 32 x}{2^{5} sin x}$
On differentiating $f\left(x\right)$ , we get
$f^{'}\left(x\right)=\frac{1}{32}\cdot \frac{32 cos 32 x sin x - cos x sin 32 x}{\left(sin\right)^{2} x}$
$\therefore f^{'}\left(\frac{\pi }{4}\right)=\frac{32 \cdot \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \cdot 0}{32 \cdot \left(\frac{1}{\sqrt{2}}\right)^{2}}=\sqrt{2}$