Difference between revisions of "Derivative of arccosh"
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(Created page with "==Theorem== The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccosh}(z) = \dfrac{1}{\sqrt{z-1}\sqrt{z+1}},$$ where $\mathrm{arccosh}$ denotes [[arccosh]...") |
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Latest revision as of 07:00, 1 January 2017
Theorem
The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccosh}(z) = \dfrac{1}{\sqrt{z-1}\sqrt{z+1}},$$ where $\mathrm{arccosh}$ denotes arccosh.
