Derivative of Arccotangent of Function

Theorem

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

Then:

$\map {\dfrac \d {\d x} } {\arccot u} = -\dfrac 1 {1 + u^2} \dfrac {\d u} {\d x}$

where $\arccot$ denotes the arccotangent of $x$.

Proof

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

\(=\)

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

Chain Rule for Derivatives

\(\ds \)

\(=\)

\(\ds -\dfrac 1 {1 + u^2} \frac {\d u} {\d x}\)

Derivative of Arccotangent Function

$\blacksquare$

Also see

• Derivative of Arcsine of Function
• Derivative of Arccosine of Function

• Derivative of Arctangent of Function

• Derivative of Arcsecant of Function
• Derivative of Arccosecant of Function

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 13$: Derivatives of Trigonometric and Inverse Trigonometric Functions: $13.23$
• 1974: Murray R. Spiegel: Theory and Problems of Advanced Calculus (SI ed.) ... (previous) ... (next): Chapter $4$. Derivatives: Derivatives of Special Functions: $16$