Derivative of Arccotangent Function

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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

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


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 defined as

This result can also be reported as:

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


Also see


Sources

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