Derivative of Inverse Hyperbolic Cotangent of x over a/Corollary

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Theorem

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

where $\size x > a$.


Proof

\(\displaystyle \dfrac 1 a \map {\coth^{-1} } {\frac x a}\) \(=\) \(\displaystyle \dfrac 1 a \cdot \dfrac 1 2 \map \ln {\dfrac {\frac x a + 1} {\frac x a - 1} }\) Definition 2 of Inverse Hyperbolic Cotangent
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 {2 a} \map \ln {\dfrac {x + a} {x - a} }\) multiplying top and bottom of argument by $a$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map {\dfrac \d {\d x} } {\dfrac 1 {2 a} \map \ln {\dfrac {x + a} {x - a} } }\) \(=\) \(\displaystyle \map {\dfrac \d {\d x} } {\dfrac 1 a \map {\coth^{-1} } {\frac x a} }\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 a \cdot \dfrac {-a} {x^2 - a^2}\) Derivative of Inverse Hyperbolic Cotangent of x over a
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 {a^2 - x^2}\) simplifying

$\blacksquare$


Sources