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Final exam - help- Inverse function Theorem

  1. Apr 30, 2012 #1
    1. The problem statement, all variables and given/known data
    Let f(x) = sinh(x) and let g be the inverse function of f. Using inverse function theorem, obtain g'(y) explicitly, a formula in y.

    Okay the Inverse function Theorem says (f^-1)'(y) = 1/(f'(x))
    If f is continuous on [a, b} and differentiable with f'(x)[itex]\neq[/itex]0 for all x[itex]\in[/itex][a, b]. Then f is 1-1 and f^-1 is continuous and differentiable on f(a,b).


    2. Relevant equations



    3. The attempt at a solution
    The derivative of f(x) sinh(x) is cosh = (e^x + e^-x)/2

    How do I get g'(y) as a formula in y? Right now, I have 2/(e^x + e^-x)
     
    Last edited by a moderator: Apr 30, 2012
  2. jcsd
  3. Apr 30, 2012 #2

    LCKurtz

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    You have ##\frac 1 {f'(x)} = \frac 1 {\cosh x}##, and you have ##y=\sinh x##. It is almost always a mistake to put in exponentials. Use the basic sinh and cosh identity to express ##\cosh x## in terms of ##y##, given what you have.
     
  4. Apr 30, 2012 #3
    Thanks so much, you save me from messing with exponents! Since I know cosh (x) = √(1 + sinh(x)^2 = √1 + y^2 !
     
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