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Hyperbolic help

  1. Dec 3, 2003 #1
    at what point on the curve [tex]y=\cosh x[/tex] does the tangent have slope 1

    I have no idea how to approach this problem

    my work

    [tex]1=\sinh x\frac{dy}{dx}[/tex]

    [tex]\frac{1}{sinh x}=\frac{dy}{dx}[/tex]
  2. jcsd
  3. Dec 3, 2003 #2


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    Science Advisor
    Homework Helper

    It would probably help to you use the definitions.

    For example:
    [tex]\sinh{x}= \frac{e^{x}-e^{-x}}{2}[/tex]
  4. Dec 3, 2003 #3
    yah I found the answer as x=ln[1+sqt2]----thanks for your help
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