Difference between revisions of "Derivative of coth"

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==Proof==
 
==Proof==
 +
By the definition,
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$$\mathrm{coth}(z) = \dfrac{\mathrm{cosh}(z)}{\mathrm{sinh}(z)}.$$
 +
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
 +
$$\begin{array}{ll}
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\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{coth}(z) &= \dfrac{\sinh^2(z)-\cosh^2(z)}{\mathrm{sinh}^2(z)} \\
 +
&= - \dfrac{\mathrm{cosh}^2(z)-\mathrm{sinh}^2(z)}{\mathrm{sinh}^2(z)} \\
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&= -\mathrm{csch}^2(z),
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\end{array}$$
 +
as was to be shown.
  
 
==References==
 
==References==
  
 
[[Category:Theorem]]
 
[[Category:Theorem]]
[[Category:Unproven]]
+
[[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.

References