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Inverse and original function relationships

  1. Jan 8, 2014 #1
    Just curious:

    Are there any unique relationships b/w the inverse of a function and the original, specifically when considering the derivative and integral?
     
  2. jcsd
  3. Jan 8, 2014 #2
    wiki for derivative and Wolfram for the integral. Is that what you're looking for?
     
  4. Jan 9, 2014 #3

    HallsofIvy

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    Use the chain rule on [itex]f(f^{-1}(x))[/itex] to get the first and "integration by parts" to get the second.
     
  5. Jan 9, 2014 #4
    Thank you for the link.



    Just wondering...if you have an original function ##f(x) = y## and the inverse, ##g(y) = x = f^{-1} (f(x))##.

    Then differentiating ##g(y)## wrt to x you get:

    ## \frac {d(g(y)}{dy} \frac {dy}{dx} = g'(y)y' = g'(f(x))f'(x) = f'^{-1} (f(x)) f'(x)##

    I'm just slightly confused on how the article derived ## f'^{-1} (x) = \frac {1}{f' ( f^{-1} (x))}## since all it says is that chain rule is used, specifically here: http://en.wikipedia.org/wiki/Inverse_functions_and_differentiation#Additional_properties.

    Also, just a little lower in the article, it states:

    ##\frac {d^2y}{dx^2} . \frac {dx}{dy} + \frac {d^2x}{dy^2} . (\frac {dy}{dx})^2 = 0##

    All I can really simplify this to by simple manipulation is:

    ##\frac {(d^2y)(dy^2)}{(d^2x)(dx^2)} = -(\frac {dy}{dx})^3##

    This isn't really helpful and I don't exactly know what else to do from here. Any insight on taking a different approach or how to move on would be greatly appreciated. I feel like I'm just not applying chain rule correctly, but any help would be great!
     
    Last edited: Jan 9, 2014
  6. Jan 9, 2014 #5
    Just use ##g(f(x))=x## and differentiate. I think it actually works faster if you use ##f(g(x))=x## like HoI, but both work fine.
     
  7. Jan 10, 2014 #6
    Hmm....I must say that my understanding of Liebniz notation is not too strong.

    For example, what exactly does this represent (if ##y = f(x)##):

    ##\frac {d^3y}{dx^3}##

    Is it simply ##\frac {1}{f'''^{-1}(x)}## ?



    Also, is ## \frac {d^2y}{dx^2} . \frac {dx}{dy} + \frac {d^2x}{dy^2} . (\frac {dy}{dx})^3 = 0 ##

    equal to ## \frac {1}{f''^{-1}(x)}. \frac {1}{f^{-1}} + \frac {1}{f''(x)}. (f'(x))^3 = 0 ##

    Besides rewriting it, I am still a little confused on how this was simplified in the next steps in the link.
     
    Last edited: Jan 10, 2014
  8. Jan 10, 2014 #7
    I am also going through the expression:

    ##f^{-1}(x) = \int \frac {1}{f'(f^{-1}(x))}dx + c##

    By taking the derivative of both sides, what I simplify this to is:

    ##f'^{-1}(x) = \frac {1}{f'(f^{-1}(x))}##

    I just don't exactly see how this result was derived using Chain Rule in the link.
     
  9. Jan 15, 2014 #8
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