Difference between revisions of "Derivative of sinh"
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| − | + | ==Theorem== | |
| − | + | The following formula holds: | |
| − | $$\dfrac{\mathrm{d}}{\mathrm{d} | + | $$\dfrac{\mathrm{d}}{\mathrm{d}z} \sinh(z) = \cosh(z),$$ |
where $\sinh$ denotes the [[sinh|hyperbolic sine]] and $\cosh$ denotes the [[cosh|hyperbolic cosine]]. | where $\sinh$ denotes the [[sinh|hyperbolic sine]] and $\cosh$ denotes the [[cosh|hyperbolic cosine]]. | ||
| − | + | ||
| − | + | ==Proof== | |
| − | + | From the definition, | |
| − | + | $$\sinh(z) = \dfrac{e^z-e^{-z}}{2},$$ | |
| + | and so using the [[derivative of the exponential function]], the [[derivative is a linear operator|linear property of the derivative]], the [[chain rule]], and the definition of the hyperbolic cosine, | ||
| + | $$\dfrac{\mathrm{d}}{\mathrm{d}z} \sinh(z)=\dfrac{e^z + e^{-z}}{2}=\cosh(z),$$ | ||
| + | as was to be shown. █ | ||
| + | |||
| + | ==References== | ||
| + | |||
| + | [[Category:Theorem]] | ||
| + | [[Category:Proven]] | ||
Latest revision as of 07:52, 8 June 2016
Theorem
The following formula holds: $$\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. █
