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