Derivative of Cotangent Function

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Theorem

$\displaystyle D_x \left({\cot x}\right) = -\csc^2 x = \frac {-1} {\sin^2 x}$, when $\sin x \ne 0$.


Proof


Then:

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