Derivative of Hyperbolic Secant/Proof 2
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Theorem
- $\map {\dfrac \d {\d x} } {\sech x} = -\sech x \tanh x$
Proof
\(\ds \map {\frac \d {\d x} } {\sech x}\) | \(=\) | \(\ds 2 \map {\frac \d {\d x} } {\frac {e^x} {e^{2 x} + 1} }\) | 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 |
$\blacksquare$