Derivative of Cotangent of Function

Theorem

Let $u$ be a differentiable real function of $x$.

Then:

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

where $\cot$ is the cotangent function and $\csc$ is the cosecant function.

Proof

\(\ds \map {\frac \d {\d x} } {\cot u}\)

\(=\)

\(\ds \map {\frac \d {\d u} } {\cot u} \frac {\d u} {\d x}\)

Chain Rule for Derivatives

\(\ds \)

\(=\)

\(\ds -\csc^2 u \frac {\d u} {\d x}\)

Derivative of Cotangent Function

$\blacksquare$

Also see

• Derivative of Sine of Function
• Derivative of Cosine of Function

• Derivative of Tangent of Function

• Derivative of Secant of Function
• Derivative of Cosecant of Function

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 13$: Derivatives of Trigonometric and Inverse Trigonometric Functions: $13.17$
• 1974: Murray R. Spiegel: Theory and Problems of Advanced Calculus (SI ed.) ... (previous) ... (next): Chapter $4$. Derivatives: Derivatives of Special Functions: $6$