Derivative of Inverse Hyperbolic Tangent Function
Theorem
Let $u$ be a differentiable real function of $x$.
Then:
- $\map {\dfrac \d {\d x} } {\artanh u} = \dfrac 1 {1 - u^2} \dfrac {\d u} {\d x}$
where $\size u < 1$
where $\tanh^{-1}$ is the real area hyperbolic tangent.
Proof
| \(\ds \map {\frac \d {\d x} } {\artanh u}\) | \(=\) | \(\ds \map {\frac \d {\d u} } {\artanh u} \frac {\d u} {\d x}\) | Chain Rule for Derivatives | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 1 {1 - u^2} \frac {\d u} {\d x}\) | Derivative of Inverse Hyperbolic Tangent |
$\blacksquare$
Also see
- Derivative of Inverse Hyperbolic Sine Function
- Derivative of Real Area Hyperbolic Cosine of Function
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 13$: Derivatives of Hyperbolic and Inverse Hyperbolic Functions: $13.39$
- 1972: Frank Ayres, Jr. and J.C. Ault: Theory and Problems of Differential and Integral Calculus (SI ed.) ... (previous) ... (next): Chapter $15$: Differentiation of Hyperbolic Functions: Definitions of Inverse Hyperbolic Functions: $39$
- 1974: Murray R. Spiegel: Theory and Problems of Advanced Calculus (SI ed.) ... (previous) ... (next): Chapter $4$. Derivatives: Derivatives of Special Functions: $27$