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

$\blacksquare$