Difference between revisions of "Derivative of hyperbolic cosecant"

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==Proof==
 
==Proof==
 
+
From the definition,
 +
$$\dfrac{\mathrm{csch}(z)} = \dfrac{1}{\mathrm{sinh}(z)}.$$
 +
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==
 
==References==
  
 
[[Category:Theorem]]
 
[[Category:Theorem]]
 
[[Category:Unproven]]
 
[[Category:Unproven]]

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, $$\dfrac{\mathrm{csch}(z)} = \dfrac{1}{\mathrm{sinh}(z)}.$$ 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