Difference between revisions of "Derivative of sinh"
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<strong>[[Derivative of sinh|Proposition]]:</strong> The following formula holds: | <strong>[[Derivative of sinh|Proposition]]:</strong> The following formula holds: | ||
| − | $$\dfrac{\mathrm{d}}{\mathrm{d} | + | $$\dfrac{\mathrm{d}}{\mathrm{d}z} \sinh(z) = \cosh(z),$$ |
where $\sinh$ denotes the [[sinh|hyperbolic sine]] and $\cosh$ denotes the [[cosh|hyperbolic cosine]]. | where $\sinh$ denotes the [[sinh|hyperbolic sine]] and $\cosh$ denotes the [[cosh|hyperbolic cosine]]. | ||
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Revision as of 00:24, 11 May 2016
Proposition: The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \sinh(z) = \cosh(z),$$ where $\sinh$ denotes the hyperbolic sine and $\cosh$ denotes the hyperbolic cosine.
Proof: From the definition, $$\sinh(z) = \dfrac{e^z-e^{-z}}{2},$$ and so using the derivative of the exponential function, the linear property of the derivative, the chain rule, and the definition of the hyperbolic cosine, $$\dfrac{\mathrm{d}}{\mathrm{d}z} \sinh(z)=\dfrac{e^z + e^{-z}}{2}=\cosh(z),$$ as was to be shown. █
