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

• 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: $13.39$
• 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: $27$