Difference between revisions of "Derivative of hyperbolic cosecant"

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==Theorem==
<strong>Proposition:</strong> $\dfrac{d}{dx}$[[Csch|$\mathrm{csch}$]]$(x)=-\mathrm{csch}(x)$[[Coth|$\mathrm{coth}$]]$(x)$
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The following formula holds:
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$$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{csch}(z)=-\mathrm{csch}(z)\mathrm{coth}(z),$$
<strong>Proof:</strong> █
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where $\mathrm{csch}$ denotes the [[csch|hyperbolic cosecant]] and $\mathrm{coth}$ denotes the [[coth|hyperbolic cotangent]].
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==Proof==
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From the definition,
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$$\mathrm{csch}(z) = \dfrac{1}{\mathrm{sinh}(z)}.$$
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Using the [[quotient rule]], the [[derivative of sinh]], and the definition of $\mathrm{coth}$, we compute
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$$\begin{array}{ll}
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\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{csch}(z) &= \dfrac{0-\mathrm{cosh}(z)}{\mathrm{sinh}^2(z)} \\
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&= -\mathrm{csch}(z)\mathrm{coth}(z),
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\end{array}$$
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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 12:13, 17 September 2016

Theorem

The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{csch}(z)=-\mathrm{csch}(z)\mathrm{coth}(z),$$ where $\mathrm{csch}$ denotes the hyperbolic cosecant and $\mathrm{coth}$ denotes the hyperbolic cotangent.

Proof

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

References