Derivative of Cotangent Function/Proof
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Theorem
- $\map {\dfrac \d {\d x} } {\cot x} = -\csc^2 x = \dfrac {-1} {\sin^2 x}$
where $\sin x \ne 0$.
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.
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Differentiation: Quotient