Derivative of Cotangent Function

Theorem

 * $\map {\dfrac \d {\d x} {\cot x} = -\csc^2 x = \dfrac {-1} {\sin^2 x}$

where $\sin x \ne 0$.

Proof
From the definition of the cotangent function:
 * $\cot x = \dfrac {\cos x} {\sin x}$

From Derivative of Sine Function:
 * $\map {\dfrac \d {\d x} {\sin x} = \cos x$

From Derivative of Cosine Function:
 * $\map {\dfrac \d {\d x} {\cos x}= -\sin x$

Then:

{{eqn | l = \map {\dfrac \d {\d x} {\cot x}     | r = \frac {\map \sin x {-\sin x} - \cos x \cos x} {\sin^2 x}      | c = Quotient Rule for Derivatives }}

This is valid only when $\sin x \ne 0$.

The result follows from the definition of the real cosecant function.

Also see

 * Derivative of Sine Function
 * Derivative of Cosine Function


 * Derivative of Tangent Function


 * Derivative of Secant Function
 * Derivative of Cosecant Function