Question
Prove that the function f given by f (x) = log |cos x| is decreasing on $\left(0, \frac{\pi}{2}\right)$ and increasing on $\left(\frac{3 \pi}{2}, 2 \pi\right)$.

Answer

It is given that f(x) = log |cos x|
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\cos \mathrm{x}}(-\sin \mathrm{x})=-\tan \mathrm{x}$
In interval $\left(0, \frac{\pi}{2}\right), f^{\prime}(x)=-\tan x<0$ 
Therefore, f is strictly decreasing on $\left(0, \frac{\pi}{2}\right)$
In interval $\left(\frac{3 \pi}{2}, 2 \pi\right), f^{\prime}(x)=-\tan x>0$ 
Therefore, f is strictly increasing in $\left(\frac{3 \pi}{2}, 2 \pi\right)$.

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