Derivative of Cosecant Function/Proof 2
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Theorem
- $\map {\dfrac \d {\d x} } {\csc x} = -\csc x \cot x$
where $\sin x \ne 0$.
Proof
| \(\ds \map {\dfrac \d {\d x} } {\csc x}\) | \(=\) | \(\ds \map {\dfrac \d {\d x} } {\dfrac 1 {\sin x} }\) | Definition of Real Cosecant Function | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {\sin x \map {\frac \d {\d x} } 1 - 1 \map {\frac \d {\d x} }{\sin x} } {\sin^2 x}\) | Quotient Rule for Derivatives | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac {0 - \cos x} {\sin^2 x}\) | Derivative of Sine Function, Derivative of Constant | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\csc x \cot x\) | Definition of Real Cosecant Function, Definition of Real Cotangent Function |
This is valid only when $\sin x \ne 0$.
$\blacksquare$
Proof
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Differentiation: Quotient