# Derivative of Inverse Hyperbolic Tangent of x over a/Corollary

## 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

$\blacksquare$