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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Revision as of 00:22, 16 September 2016

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.

Proof

References