Derivative of Arccotangent Function/Proof 2
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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 |
$\blacksquare$