Derivative of Hyperbolic Secant

From ProofWiki
Jump to navigation Jump to search

Theorem

$\map {\dfrac \d {\d x} } {\sech x} = -\sech x \tanh x$

where $\tanh$ is the hyperbolic tangent and $\sech$ is the hyperbolic secant.


Proof 1

\(\ds \map {\frac \d {\d x} } {\sech x}\) \(=\) \(\ds \map {\frac \d {\d x} } {\frac 1 {\cosh x} }\) Definition of Hyperbolic Secant
\(\ds \) \(=\) \(\ds \map {\frac \d {\d x} } {\paren {\cosh x}^{-1} }\) Exponent Laws
\(\ds \) \(=\) \(\ds -\paren {\cosh x}^{-2} \sinh x\) Derivative of Hyperbolic Cosine, Power Rule for Derivatives, Chain Rule for Derivatives
\(\ds \) \(=\) \(\ds \frac {-1} {\cosh x} \frac {\sinh x} {\cosh x}\) Exponent Combination Laws
\(\ds \) \(=\) \(\ds -\sech z \tanh z\) Definition of Hyperbolic Secant and Definition of Hyperbolic Tangent

$\blacksquare$


Proof 2

\(\ds \map {\frac \d {\d x} } {\sech x}\) \(=\) \(\ds 2 \map {\frac \d {\d x} } {\frac {e^x} {e^{2 x} + 1} }\) Definition of Hyperbolic Secant
\(\ds \) \(=\) \(\ds \frac 2 {\paren {e^{2 x} + 1}^2} \paren {\map {\frac \d {\d x} } {e^x} \paren {e^{2 x} + 1} - e^x \map {\frac \d {\d x} } {e^{2 x} + 1} }\) Quotient Rule for Derivatives
\(\ds \) \(=\) \(\ds -\frac 2 {\paren {e^{2 x} + 1}^2} \paren {2 e^{2 x} \cdot e^x - e^x \cdot e^{2 x} - e^x}\) Derivative of Exponential Function
\(\ds \) \(=\) \(\ds -\frac {2 \paren {e^{3 x} - e^x} } {\paren {e^{2 x} + 1}^2}\)
\(\ds \) \(=\) \(\ds -\frac {2 e^x} {\paren {e^{2 x} + 1} } \cdot \frac {e^{2 x} - 1} {e^{2 x} + 1}\)
\(\ds \) \(=\) \(\ds -\sech x \tanh x\) Definition of Hyperbolic Secant, Definition of Hyperbolic Tangent

$\blacksquare$


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Derivative_of_Hyperbolic_Secant&oldid=543619"