Difference between revisions of "Derivative of hyperbolic cosecant"
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Using the [[quotient rule]], the [[derivative of sinh]], and the definition of $\mathrm{coth}$, we compute | Using the [[quotient rule]], the [[derivative of sinh]], and the definition of $\mathrm{coth}$, we compute | ||
$$\begin{array}{ll} | $$\begin{array}{ll} | ||
| − | \dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{csch}(z) = \dfrac{0-\mathrm{cosh}(z)}{\mathrm{sinh}^2(z)} \\ | + | \dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{csch}(z) &= \dfrac{0-\mathrm{cosh}(z)}{\mathrm{sinh}^2(z)} \\ |
&= -\mathrm{csch}(z)\mathrm{coth}(z), | &= -\mathrm{csch}(z)\mathrm{coth}(z), | ||
\end{array}$$ | \end{array}$$ | ||
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[[Category:Theorem]] | [[Category:Theorem]] | ||
| − | [[Category: | + | [[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.
