We know that the derivative of $\cot x$ with respect to $x$ is $-\csc^2 x$. Then, since $-\csc^2 x<0$ is negative except for integer multiples of $\pi$, therefore, can we claim that the function $\cot x$ is decreasing over the interval $(-\frac{\pi}2,\frac{\pi}2)$ except for the point $x=0$ for which the function is not defined?
Best Answer
No. A function $f$ is decreasing on $D \subseteq \mathbb{R}$ if for every $x$ and $y$ in $D$, we have that $x≤y \Rightarrow f(x)≥f(y)$.
So for $D = \left( - \frac{π}{2}, 0\right)\cup \left(0, \frac{π}{2}\right) $, choosing $x \in \left(-\frac{π}{2}, 0\right)$ and $y \in \left(0, \frac{π}{2} \right)$ with $f(x)= \cot x$, you'd have, $x≤y \Rightarrow f(x)≤f(y)$.
So does that mean $f$ is increasing? No. Choose $x$ and $y$ from $\left(-\frac{π}{2},0\right)$ and see for yourself.
Well then what in the world is it? Well $f$ is a neither increasing nor decreasing (or non-monotonic) function on $D$.
There's a good lesson to learn from this— that it is meaningless to say a function is increasing or decreasing without mentioning where. For instance, our $f$ is indeed decreasing on the intervals $ \left( - \frac{π}{2}, 0\right)$ and $ \left(0, \frac{π}{2}\right)$ but just not on both of them together.