If \[f(x)=\cos x\cos 2x\cos 4x\cos 8x\cos 16x\], then \[{f}'\left( \frac{\pi }{4} \right)\] is [AMU 2005]

If \[f(x)=\cos x\cos 2x\cos 4x\cos 8x\cos 16x\], then \[{f}'\left( \fr

1.	If \[f(x)=\cos x\cos 2x\cos 4x\cos 8x\cos 16x\], then \[{f}'\left( \frac{\pi }{4} \right)\] is [AMU 2005]
A.	\[\sqrt{2}\]
B.	\[\frac{1}{\sqrt{2}}\]
C.	1
D.	\[\frac{\sqrt{3}}{2}\]
Answer» B. \[\frac{1}{\sqrt{2}}\]

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If \[f(x)=\cos x\cos 2x\cos 4x\cos 8x\cos 16x\], then \[{f}'\left( \frac{\pi }{4} \right)\] is [AMU 2005]

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