Derivative of Hyperbolic Tangent/Corollary
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Theorem
- $\map {\dfrac \d {\d x} } {\tanh x} = 1 - \tanh^2 x$
where $\tanh$ is the hyperbolic tangent.
Proof
\(\ds \map {\dfrac \d {\d x} } {\tanh x}\) | \(=\) | \(\ds \sech^2 x\) | Derivative of Hyperbolic Tangent | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 - \tanh^2 x\) | Sum of Squares of Hyperbolic Secant and Tangent |
$\blacksquare$
Sources
- 1976: K. Weltner and W.J. Weber: Mathematics for Engineers and Scientists ... (previous) ... (next): $5$. Differential Calculus: Appendix: Derivatives of fundamental functions: $6.$ Hyperbolic trigonometric functions