Derivatives of cotangent function

Derivative of cot:
Based on the above discussion we see that cot function exists (or is defined) only if sin(x) at that point is not equal to zero. Therefore for the derivative of cot function to exist, the following condition has to be met:
 Interval (a,b) should not contain n(pi) for any n Є Z, (Z = set of all integers). That is because, for x = n(pi), sin (x) = sin (n(pi)) = 0. And we already established that sin function cannot be equal to zero for cot function to exist at that point.

Derivative of cot x:
If (a,b) does not contain 

Assuming that sin and cos are differentiable and x ≠ n(pi), then sin x ≠ 0.


Proof for derivative of cotx:

The above derivation was using the quotient rule. We can also obtain the above form using the limit definition of derivative as follows:
Derivative of cot x (using limit definition):

Derivative of cot(-x):

(using chain rule, therefore multiplying by (d/dx) (-x) = -1)

= csc^2(x)