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 Real Area Hyperbolic Cosine of Function

• Derivative of Inverse Hyperbolic Tangent Function
• Derivative of Inverse Hyperbolic Cotangent Function

• Derivative of Inverse Hyperbolic Secant Function
• Derivative of Inverse Hyperbolic Cosecant 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: $25$