Difference between revisions of "Derivative of arcsinh"

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==Theorem==
<strong>[[Derivative of arcsinh|Theorem]]:</strong> The following formula holds:
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The following formula holds:
 
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arcsinh}(z) = \dfrac{1}{\sqrt{1+z^2}},$$
 
$$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arcsinh}(z) = \dfrac{1}{\sqrt{1+z^2}},$$
 
where $\mathrm{arcsinh}$ denotes the [[arcsinh|inverse hyperbolic sine]].
 
where $\mathrm{arcsinh}$ denotes the [[arcsinh|inverse hyperbolic sine]].
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<strong>Proof:</strong> █
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==Proof==
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==References==
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[[Category:Theorem]]
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[[Category:Unproven]]

Revision as of 07:57, 8 June 2016

Theorem

The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arcsinh}(z) = \dfrac{1}{\sqrt{1+z^2}},$$ where $\mathrm{arcsinh}$ denotes the inverse hyperbolic sine.

Proof

References