Derivative of Cosecant Function/Proof 1
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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:
\(\ds \map {\dfrac \d {\d x} } {\csc x}\) | \(=\) | \(\ds \cos x \frac {-1} {\sin^2 x}\) | Chain Rule for Derivatives | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {-1} {\sin x} \frac {\cos x} {\sin x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\csc x \cot x\) | Definition of Real Cosecant Function and Definition of Real Cotangent Function |
This is valid only when $\sin x \ne 0$.
$\blacksquare$