If f(x)=cosx.cos2x.cos4x.cos8x.cos16x, then f'(π4) is

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

1. A

$\sqrt{2}$

2. B

$\frac{1}{\sqrt{2}}$

3. C

1

4. D

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### Solution:

$f\left(x\right)=\mathrm{cos}x.\mathrm{cos}2x.\mathrm{cos}4x.\mathrm{cos}8x.\mathrm{cos}16x$

$\mathrm{log}f\left(x\right)=\mathrm{logcos}x+\mathrm{logcos}2x+\cdots +\mathrm{logcos}16x$

$\frac{f\text{'}\left(x\right)}{f\left(x\right)}=-\frac{\mathrm{sin}x}{\mathrm{cos}x}-\frac{2\mathrm{sin}2x}{\mathrm{cos}2x}-\cdots -\frac{16\mathrm{sin}16x}{\mathrm{cos}16x}$

$f\text{'}\left(x\right)=-f\left(x\right)\left[\mathrm{tan}x+2\mathrm{tan}x+\cdots +16\mathrm{tan}16x\right]$

$\text{at\hspace{0.17em}\hspace{0.17em}}x=\frac{\pi }{4}$

$f\text{'}\left(\frac{\pi }{4}\right)=0$

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