Derivative of Arccotangent Function/Proof 2

Theorem

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


Proof

\(\ds \frac {\map \d {\arccot x} } {\d x}\) \(=\) \(\ds \map {\frac \d {\d x} } {\frac \pi 2 - \arctan x}\) Tangent of Complement equals Cotangent
\(\ds \) \(=\) \(\ds -\frac 1 {1 + x^2}\) Derivative of Arctangent Function

$\blacksquare$

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