Where are the critical points of $cot x$ $cot x$ ?

Question

4173 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-11-07T15:01:42

Let $f (x) = cot x = \frac{cos x}{sin x}$ $f(x)=cotx={cosx}/{sinx}$ .

By taking the derivative,

$f'(x)=-csc^2x=-1/{sin^2x} ne0$

and

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

Hence, there is no critical point.

I hope that this was helpful.

Where are the critical points of cot xcotxcot x?