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The Reciprocal of a Derivative

  1. Feb 28, 2010 #1
    I have seen this sort of thing over the past few years and it is bothering me, something like this


    But it seems to me that this sort of thing only works in certain situations. For example, take s(x) to be


    so that


    Now to get the inversion, use [tex]x=\pm\sqrt{s}[/tex]


    Now, we know that x is +/- sqrt(s), and I suppose that you could look at it like this: when x is positive, its value is given by +sqrt(s) and when x is negative it given by -sqrt(s) and so we can just directly replace +/- sqrt(s) with x.



    But, if you back it up a bit and instead substitute [tex]s=x^{2}[/tex] into


    to get


    which results in


    because x squared is always positive. Then invert it to get

    [tex]\frac{1}{\frac{dx}{ds}}=\pm 2x[/tex]

    Which doesn't agree with the earlier method. So, depending on how you look at it, it seems that you get different answers. However, this isn't a problem if s is bijective. So, does this 'only' work when the function is bijective? Or, possibly, there is something wrong with my way of viewing these relationships and the +/- isn't there in the final result? What do you think about this inversion in general?
    Last edited: Feb 28, 2010
  2. jcsd
  3. Feb 28, 2010 #2
    This would also equal to 1/(2x). The square root takes the principle root. Thus the range of Sqr(x2) does not cover the original domain of x. The +/- Sqr(x2) covers for that.
    Last edited: Feb 28, 2010
  4. Feb 28, 2010 #3
    So it should be something more like:


    That takes care of the case that I brought up but does it always work?
  5. Feb 28, 2010 #4
  6. Feb 28, 2010 #5
    Holy crap, thanks for the link.
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