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