Difference between revisions of "Derivative of sinh"
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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]]. | ||
<div class="mw-collapsible-content"> | <div class="mw-collapsible-content"> | ||
| − | <strong>Proof:</strong> █ | + | <strong>Proof:</strong> 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. █ | ||
</div> | </div> | ||
</div> | </div> | ||
Revision as of 04:49, 7 May 2016
Proposition: The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}x} \sinh(x) = \cosh(x),$$ 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. █
