Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook