Derivative of Cotangent Function

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Theorem

$D_x \left({\cot x}\right) = -\csc^2 x = \dfrac {-1} {\sin^2 x}$

where $\sin x \ne 0$.


Corollary

$D_x \left({\cot a x}\right) = -a \csc^2 a x$


Proof

From the definition of the cotangent function:

$\cot x = \dfrac {\cos x} {\sin x}$

From Derivative of Sine Function:

$D_x \left({\sin x}\right) = \cos x$

From Derivative of Cosine Function:

$D_x \left({\cos x}\right) = -\sin x$


Then:

\(\displaystyle D_x \left({\cot x}\right)\) \(=\) \(\displaystyle \frac {\sin x \left({-\sin x}\right) - \cos x \cos x} {\sin^2 x}\)          Quotient Rule for Derivatives          
\(\displaystyle \) \(=\) \(\displaystyle \frac {-\left({\sin^2 x + \cos^2 x}\right)} {\sin^2 x}\)                    
\(\displaystyle \) \(=\) \(\displaystyle \frac {-1} {\sin^2 x}\)          Sum of Squares of Sine and Cosine          

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

The result follows from the definition of the cosecant function.

$\blacksquare$


Sources