Derivative of Inverse Hyperbolic Cotangent of x over a

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Theorem

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

where $x^2 > a^2$.


Corollary

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

where $\size x > a$.


Proof

Let $x^2 > a^2$.

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

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


$\coth^{-1} \dfrac x a$ is not defined when $x^2 \le a^2$.

$\blacksquare$


Also presented as

This result can also be (and usually is) reported as:

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

but this obscures the fact that $x^2 > a^2$ in order for it to be defined.

Some sources present it as:

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


Also see


Sources