# Where are the critical points of cot x?

Nov 7, 2014

Let $f \left(x\right) = \cot x = \frac{\cos x}{\sin x}$.

By taking the derivative,

$f ' \left(x\right) = - {\csc}^{2} x = - \frac{1}{{\sin}^{2} x} \ne 0$

and

$f '$ is always defined in the domain of $f$.

Hence, there is no critical point.

I hope that this was helpful.