Difference between revisions of "Derivative of coth"
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| − | + | ==Theorem== | |
| − | + | The following formula holds: | |
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{coth}(z) = -\mathrm{csch}^2(z),$$ | $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{coth}(z) = -\mathrm{csch}^2(z),$$ | ||
where $\mathrm{coth}$ denotes the [[coth|hyperbolic cotangent]] and $\mathrm{csch}$ denotes the [[csch|hyperbolic cosecant]]. | where $\mathrm{coth}$ denotes the [[coth|hyperbolic cotangent]] and $\mathrm{csch}$ denotes the [[csch|hyperbolic cosecant]]. | ||
| − | + | ||
| − | + | ==Proof== | |
| − | + | By the definition, | |
| − | + | $$\mathrm{coth}(z) = \dfrac{\mathrm{cosh}(z)}{\mathrm{sinh}(z)}.$$ | |
| + | Using the [[quotient rule]], the [[derivative of sinh]], the [[derivative of cosh]], [[Pythagorean identity for sinh and cosh]], and the definition of $\mathrm{csch}$, we see | ||
| + | $$\begin{array}{ll} | ||
| + | \dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{coth}(z) &= \dfrac{\sinh^2(z)-\cosh^2(z)}{\mathrm{sinh}^2(z)} \\ | ||
| + | &= - \dfrac{\mathrm{cosh}^2(z)-\mathrm{sinh}^2(z)}{\mathrm{sinh}^2(z)} \\ | ||
| + | &= -\mathrm{csch}^2(z), | ||
| + | \end{array}$$ | ||
| + | as was to be shown. | ||
| + | |||
| + | ==References== | ||
| + | |||
| + | [[Category:Theorem]] | ||
| + | [[Category:Proven]] | ||
Latest revision as of 12:18, 17 September 2016
Theorem
The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{coth}(z) = -\mathrm{csch}^2(z),$$ where $\mathrm{coth}$ denotes the hyperbolic cotangent and $\mathrm{csch}$ denotes the hyperbolic cosecant.
Proof
By the definition, $$\mathrm{coth}(z) = \dfrac{\mathrm{cosh}(z)}{\mathrm{sinh}(z)}.$$ Using the quotient rule, the derivative of sinh, the derivative of cosh, Pythagorean identity for sinh and cosh, and the definition of $\mathrm{csch}$, we see $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{coth}(z) &= \dfrac{\sinh^2(z)-\cosh^2(z)}{\mathrm{sinh}^2(z)} \\ &= - \dfrac{\mathrm{cosh}^2(z)-\mathrm{sinh}^2(z)}{\mathrm{sinh}^2(z)} \\ &= -\mathrm{csch}^2(z), \end{array}$$ as was to be shown.
