If $f\left(x\right)=\cos x·\cos 2x·\cos 4x·\cos 8x·\cos 16x$ then $\mathrm{f}\text{'}\left(\frac{\mathrm{\pi }}{4}\right)$ is 

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=\cos \mathrm{x} \cos 2\mathrm{x} \cos 4\mathrm{x} \cos 8\mathrm{x} \cos 16\mathrm{x} \\ \mathrm{f}\left(\mathrm{x}\right)=\frac{\sin 32\mathrm{x}}{32 \sin \mathrm{x}} \\ {\mathrm{f}}^1\left(\mathrm{x}\right)=\frac{\sin \mathrm{x} 32 \cos 32\mathrm{x}-\cos \mathrm{x} \sin 32\mathrm{x}}{32 {\sin }^2\mathrm{x}} \\ {\mathrm{f}}^1\left(\frac{\mathrm{\pi }}{4}\right)=\frac{{\displaystyle \frac{1}{\sqrt{2}}}32\left(1\right)-0}{32\left({\displaystyle \frac{1}{2}}\right)} \\ =\sqrt{2} \end{gathered}$$