Question

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

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

Answer 1

Answer: (A)

We have, $f\left(x\right)=cosxcos2xcos4xcos8xcos16x$ $\Rightarrow f\left(x\right)=\frac{2sinxcosxcos2xcos4xcos8xcos16x}{2sinx}$
$=\frac{sin32x}{2^{5}sinx}$
On differentiating $f\left(x\right)$ , we get
$f^{{}^{\prime }}\left(x\right)=\frac{1}{32}\cdot \frac{32cos32xsinx-cosxsin32x}{{\left(sin\right)}^{2}x}$
$\therefore f^{{}^{\prime }}\left(\frac{\pi }{4}\right)=\frac{32\cdot \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\cdot 0}{32\cdot {\left(\frac{1}{\sqrt{2}}\right)}^2}=\sqrt{2}$