Derivative of Arccotangent Function

Theorem

Let $x \in \R$.

Let $\arccot x$ be the arccotangent of $x$.

Then:

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

Corollary

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

Proof 1

$\ds y$

$=$

$\ds \arccot x$

$\ds \leadsto \ \ $

$\ds x$

$=$

$\ds \cot y$

Definition of Real Arccotangent

$\ds \leadsto \ \ $

$\ds \frac {\d x} {\d y}$

$=$

$\ds -\csc^2 y$

Derivative of Cotangent Function

$\ds $

$=$

$\ds -\paren {1 + \cot^2 y}$

Difference of Squares of Cosecant and Cotangent

$\ds $

$=$

$\ds -\paren {1 + x^2}$

Definition of $x$

$\ds \leadsto \ \ $

$\ds \frac {\d y} {\d x}$

$=$

$\ds \frac {-1} {1 + x^2}$

Derivative of Inverse Function

$\blacksquare$

Proof 2

$\ds \frac {\map \d {\arccot x} } {\d x}$

$=$

$\ds \map {\frac \d {\d x} } {\frac \pi 2 - \arctan x}$

Tangent of Complement equals Cotangent

$\ds $

$=$

$\ds -\frac 1 {1 + x^2}$

Derivative of Arctangent Function

$\blacksquare$

Also presented as

The can also be presented in the form:

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

Also see

Sources

1976: K. Weltner and W.J. Weber: Mathematics for Engineers and Scientists ... (previous) ... (next): $5$. Differential Calculus: Appendix: Derivatives of fundamental functions: $4.$ Inverse trigonometric functions
1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): Appendix $2$: Table of derivatives and integrals of common functions: Inverse trigonometric functions
1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): Appendix: Table $1$: Derivatives
2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Appendix: Table $1$: Derivatives