Question:
Prove that the function $f$ given by $f(x)=\log \cos x$ is strictly decreasing on $\left(0, \frac{\pi}{2}\right)$ and strictly increasing on $\left(\frac{\pi}{2}, \pi\right)$.
Solution:
We have,
$f(x)=\log \cos x$
$\therefore f^{\prime}(x)=\frac{1}{\cos x}(-\sin x)=-\tan x$
In interval $\left(0, \frac{\pi}{2}\right), \tan x>0 \Rightarrow-\tan x<0$
$\therefore f^{\prime}(x)<0$ on $\left(0, \frac{\pi}{2}\right)$
$\therefore f$ is strictly decreasing on $\left(0, \frac{\pi}{2}\right)$.
In interval $\left(\frac{\pi}{2}, \pi\right), \tan x<0 \Rightarrow-\tan x>0$
$\therefore f^{\prime}(x)>0$ on $\left(\frac{\pi}{2}, \pi\right)$
$\therefore f$ is strictly increasing on $\left(\frac{\pi}{2}, \pi\right)$.