Question

Where are the critical points of `cot x`?

Answer 1

Let `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.

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I hope that this was helpful.