Derivative of Inverse Hyperbolic Tangent of x over a/Corollary

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Theorem

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

where $\size x < a$.


Proof

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

$\blacksquare$


Sources