Derivative of Inverse Hyperbolic Sine Function
Theorem
Let $u$ be a differentiable real function of $x$.
Then:
- $\map {\dfrac \d {\d x} } {\arsinh u} = \dfrac 1 {\sqrt {1 + u^2} } \dfrac {\d u} {\d x}$
where $\sinh^{-1}$ is the inverse hyperbolic sine.
Proof
| \(\ds \map {\frac \d {\d x} } {\arsinh u}\) | \(=\) | \(\ds \map {\frac \d {\d u} } {\arsinh u} \frac {\d u} {\d x}\) | Chain Rule for Derivatives | |||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 1 {\sqrt {1 + u^2} } \frac {\d u} {\d x}\) | Derivative of Inverse Hyperbolic Sine |
$\blacksquare$
Also see
- Derivative of Inverse Hyperbolic Tangent Function
- Derivative of Inverse Hyperbolic Cotangent Function
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 13$: Derivatives of Hyperbolic and Inverse Hyperbolic Functions: $13.37$
- 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: $37$.
- 1974: Murray R. Spiegel: Theory and Problems of Advanced Calculus (SI ed.) ... (previous) ... (next): Chapter $4$. Derivatives: Derivatives of Special Functions: $25$