Derivative of Inverse Hyperbolic Secant Function

Theorem

Let $u$ be a differentiable real function of $x$.

Then:

$\map {\dfrac \d {\d x} } {\sech^{-1} u} = \dfrac {-1} {u \sqrt {1 - u^2} } \dfrac {\d u} {\d x}$

where $0 < u < 1$

where $\sech^{-1}$ is the inverse hyperbolic secant.


Proof

\(\ds \map {\frac \d {\d x} } {\sech^{-1} u}\) \(=\) \(\ds \map {\frac \d {\d u} } {\sech^{-1} u} \frac {\d u} {\d x}\) Chain Rule for Derivatives
\(\ds \) \(=\) \(\ds \dfrac {-1} {u \sqrt {1 - u^2} } \frac {\d u} {\d x}\) Derivative of Inverse Hyperbolic Secant

$\blacksquare$


Also see


Sources

  • 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 13$: Derivatives of Hyperbolic and Inverse Hyperbolic Functions: $13.41$
  • 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: $41$
  • 1974: Murray R. Spiegel: Theory and Problems of Advanced Calculus (SI ed.) ... (previous) ... (next): Chapter $4$. Derivatives: Derivatives of Special Functions: $29$