Derivative of Cosecant Function

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Theorem

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

where $\sin x \ne 0$.


Proof

From the definition of the cosecant function:

$\csc x = \dfrac 1 {\sin x}$

From Derivative of Sine Function:

$\map {\dfrac \d {\d x} } {\sin x} = \cos x$


Then:

\(\displaystyle \map {\dfrac \d {\d x} } {\csc x}\) \(=\) \(\displaystyle \cos x \frac {-1} {\sin^2 x}\) Chain Rule for Derivatives
\(\displaystyle \) \(=\) \(\displaystyle \frac {-1} {\sin x} \frac {\cos x} {\sin x}\)
\(\displaystyle \) \(=\) \(\displaystyle -\csc x \cot x\) Definition of Real Cosecant Function and Definition of Real Cotangent Function

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

$\blacksquare$


Also see