Derivative of Inverse Hyperbolic Tangent of x over a

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Theorem

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

where $-a < x < a$.


Corollary

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

where $\size x < a$.


Proof

Let $-a < x < a$.

Then $-1 < \dfrac x a < 1$ and so:

\(\displaystyle \map {\dfrac \d {\d x} } {\map {\tanh^{-1} } {\dfrac x a} }\) \(=\) \(\displaystyle \frac 1 a \frac 1 {1 - \paren {\frac x a}^2}\) Derivative of $\tanh^{-1}$ and Derivative of Function of Constant Multiple
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 a \frac 1 {\frac {a^2 - x^2} {a^2} }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 a \frac {a^2} {a^2 - x^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac a {a^2 - x^2}\)


$\tanh^{-1} \dfrac x a$ is not defined when either $x \le -a$ or $x \ge a$.

$\blacksquare$


Also presented as

Some sources present this as:

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


Also see


Sources