Derivative of Real Area Hyperbolic Tangent of x over a
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Theorem
- $\map {\dfrac \d {\d x} } {\map \artanh {\dfrac x a} } = \dfrac a {a^2 - x^2}$
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:
\(\ds \map {\dfrac \d {\d x} } {\map \artanh {\dfrac x a} }\) | \(=\) | \(\ds \frac 1 a \frac 1 {1 - \paren {\frac x a}^2}\) | Derivative of $\artanh$ and Derivative of Function of Constant Multiple | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \frac 1 {\frac {a^2 - x^2} {a^2} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \frac {a^2} {a^2 - x^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac a {a^2 - x^2}\) |
$\artanh \dfrac x a$ is not defined when either $x \le -a$ or $x \ge a$.
$\blacksquare$
Also presented as
Some sources present this as:
- $\map {\dfrac \d {\d x} } {\dfrac 1 a \map \artanh {\dfrac x a} } = \dfrac 1 {a^2 - x^2}$
Also see
Sources
- 1944: R.P. Gillespie: Integration (2nd ed.) ... (previous) ... (next): Chapter $\text {II}$: Integration of Elementary Functions: $\S 7$. Standard Integrals: $16$.