What is the derivative of cosh inverse x?

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JFonseka
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Homework Statement



Derive cosh[tex]^{-1}[/tex]x

Homework Equations



None I know of.

The Attempt at a Solution



Well I vaguely remember that the inverse of this was something like

ln(x + [tex]\sqrt{x^2 - 1}[/tex])

If I derive this, I will get [tex]\frac{1}{\sqrt{x^2 - 1}}[/tex]

Is that correct? Am I wrong to assume the equation for the inverse of cosh? Or do I need to prove that as well
 
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actually...[tex]\frac{d}{dx}arccoshx=\frac{1}{\sqrt{x^2-1}}[/tex]

prove it by just letting y=arccoshx and then putting coshy=x and findind dy/dx
and use the identity cosh^2(x)-sinh^2(x)=1