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Derivative of an inverse function

  1. Dec 6, 2009 #1
    f(x) = cosh^2(x)+sinh(2x) = y
    f'(x) = sinh(2x)+2cosh(2x) = 3e^(2x) + e^(-x) = y'

    Let g(y) be the inverse of f(x):
    g'(y) = 1 / f'(x) = 1 / [3e^(2y) + e^(-2y)] = e^(2y) / [4e^(2y) + 1]

    Integrating gives: [ 3^(1/2)/3 ]*arctan[ 3^(1/2) * e^(2y) ] + C

    Now when I plotted this function it looked in no way like the inverse of f(x), so where have I gone wrong?

    Thank you :smile:
     
  2. jcsd
  3. Dec 6, 2009 #2

    you just replaced y by x. What you have to do is use the relation y = g(x), where g is the inverse of f. Given your definition of f it might be hard to find an expression for this inverse:smile:
     
  4. Dec 7, 2009 #3
    Hello :smile:
    I don't understand why g(y) isn't the inverse?
    f: x -> y and g: y -> x
    I still don't understand why integrating 1/f'(g(y)) doesn't yield me g(y) :smile:
     
  5. Dec 8, 2009 #4

    Landau

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    This is incorrect. Although you write g'(y)=1/f'(x), you are just using g'(y)=1/f'(y).

    As said above, you should be using g'(y)=1/f'(x) where x and y are such that y=f(x)! The relation between x and y is crucial.
     
  6. Dec 8, 2009 #5
    To clarify what was said before, how is f'(x), a function of x, expressed completely in terms of y?

    You should have g'(y) = 1/[3e^(2x) + e^(-2x)] = e^(2x)/(4e^(2x) + 1)

    In which case your integration tricks won't work.
     
  7. Dec 9, 2009 #6
    That would be true. However, you seem to have tried to integrate 1/f'(y) to recover g and that does not work.
     
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