Derivative of Hyperbolic Secant Function
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Theorem
Let $u$ be a differentiable real function of $x$.
Then:
- $\map {\dfrac \d {\d x} } {\sech u} = -\sech u \tanh u \dfrac {\d u} {\d x}$
where $\tanh$ is the hyperbolic tangent and $\sech$ is the hyperbolic secant.
Proof
\(\ds \map {\frac \d {\d x} } {\sech u}\) | \(=\) | \(\ds \map {\frac \d {\d u} } {\sech u} \frac {\d u} {\d x}\) | Chain Rule for Derivatives | |||||||||||
\(\ds \) | \(=\) | \(\ds -\sech u \tanh u \frac {\d u} {\d x}\) | Derivative of 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: $13.35$
- 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: Differentiation Formulas: $35$.
- 1974: Murray R. Spiegel: Theory and Problems of Advanced Calculus (SI ed.) ... (previous) ... (next): Chapter $4$. Derivatives: Derivatives of Special Functions: $23$