Difference between revisions of "Derivative of sech"

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==Theorem==
<strong>[[Derivative of sech|Theorem]]:</strong> The following formula holds:
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The following formula holds:
 
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{sech}(z)=-\mathrm{sech}(z)\mathrm{tanh}(z),$$
 
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{sech}(z)=-\mathrm{sech}(z)\mathrm{tanh}(z),$$
 
where $\mathrm{sech}$ denotes the [[sech|hyperbolic secant]] and $\mathrm{tanh}$ denotes the [[tanh|hyperbolic tangent]].
 
where $\mathrm{sech}$ denotes the [[sech|hyperbolic secant]] and $\mathrm{tanh}$ denotes the [[tanh|hyperbolic tangent]].
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<strong>Proof:</strong> █
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==Proof==
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From the definition,
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$$\mathrm{sech}(z) = \dfrac{1}{\mathrm{cosh}(z)}.$$
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Using the [[quotient rule]], the [[derivative of cosh]], and the definition of $\mathrm{tanh}$, we see
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$$\begin{array}{ll}
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\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{sech}(z) &= \dfrac{0-\sinh(z)}{\cosh(z)^2} \\
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&=-\mathrm{sech}(z)\mathrm{tanh}(z),
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\end{array}$$
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as was to be shown.
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==References==
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[[Category:Theorem]]
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[[Category:Proven]]

Latest revision as of 11:47, 17 September 2016

Theorem

The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{sech}(z)=-\mathrm{sech}(z)\mathrm{tanh}(z),$$ where $\mathrm{sech}$ denotes the hyperbolic secant and $\mathrm{tanh}$ denotes the hyperbolic tangent.

Proof

From the definition, $$\mathrm{sech}(z) = \dfrac{1}{\mathrm{cosh}(z)}.$$ Using the quotient rule, the derivative of cosh, and the definition of $\mathrm{tanh}$, we see $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{sech}(z) &= \dfrac{0-\sinh(z)}{\cosh(z)^2} \\ &=-\mathrm{sech}(z)\mathrm{tanh}(z), \end{array}$$ as was to be shown.

References