Derivative of Cotangent Function

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Theorem

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

where $\sin x \ne 0$.


Corollary 1

$\map {\dfrac \d {\d x} } {\cot a x} = -a \csc^2 a x$


Corollary 2

$\dfrac \d {\d x} \cot x = -1 - \cot^2 x$


Corollary 3

$\map {\dfrac \d {\d x} } {\cot a x} = -a \paren {\cot^2 a x + 1}$


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:

\(\ds \map {\dfrac \d {\d x} } {\cot x}\) \(=\) \(\ds \frac {\sin x \paren {-\sin x} - \cos x \cos x} {\sin^2 x}\) Quotient Rule for Derivatives
\(\ds \) \(=\) \(\ds \frac {-\paren {\sin^2 x + \cos^2 x} } {\sin^2 x}\)
\(\ds \) \(=\) \(\ds \frac {-1} {\sin^2 x}\) Sum of Squares of Sine and Cosine

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

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

$\blacksquare$


Also see


Sources

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