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Help with a proof!

  1. Jun 28, 2012 #1
    How does one go about rigorously proving that (dy/dx)=1/(dx/dy)?

    Thanks
     
  2. jcsd
  3. Jun 28, 2012 #2
    and vice-versa of course
     
  4. Jun 28, 2012 #3
    Good question. To answer that, first ask yourself, what do you mean by [itex] \frac{dy}{dx} [/itex]? What do you mean by [itex] \frac{dx}{dy} [/itex]?

    I recommend looking at this post. https://www.physicsforums.com/showthread.php?t=63886 (it's the same as your question)
     
  5. Jun 28, 2012 #4
    how can y=f(x) define y implicitly? that seems explicit.
     
  6. Jun 28, 2012 #5
    the above is in regard to the link provided
     
  7. Jun 28, 2012 #6

    vela

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    You need to show some more effort here, per the forum rules. We're not here to do your homework for you.
     
  8. Jun 28, 2012 #7
    This isn't for homework. Here is my attempt at a proof.

    1.Say we are given a function y=f(x)

    2.Differentiate implicit w.r.t. y and we see....

    1=(df/dx)(dx/dy)

    3. Now we can solve for the derivative of our choice!

    Now, my question is.....

    Dosen't the result of 2(implicit differentiation) imply that we thought of y=f(x) as y=f(x(y))? In other words,
    implicit diff stems from the chain rule right? So, dosen't that mean that the proof above is not general, in other
    words, what if I cant solve y=f(x) for x as a function of y explicitly? Then I cant say y=f(x) is the same as y=f(x(y))
    and obtain the right side of 2 which results from the chain rule.

    Any ideas?
     
    Last edited: Jun 28, 2012
  9. Jun 28, 2012 #8
    Have you ever heard of the inverse function theorem? I will summarize it for a function of one variable, since the formulation is a bit easier. The theorem effectively says that if you have a function [itex]f:A\subset \mathbb{R}\rightarrow \mathbb{R} [/itex] which is differentiable and has a continuous derivative on some open set [itex] A [/itex], and [itex] f'(a)\neq 0 [/itex] at some point [itex] a\in A[/itex], then [itex] f^{-1} [/itex] exists and is defined in some neighbourhood of [itex] a [/itex].

    In the context of your problem, this means that as long as [itex] f [/itex] is [itex]C^1 [/itex], and its derivative isn't zero one the whole domain, then SOMEWHERE in its domain, [itex]f[/itex] is invertible. Suppose that [itex] f [/itex] is invertible on [itex] U\subset A [/itex] as above. Then we have a function [itex] f^{-1} : f(U)\rightarrow A [/itex] which satisfies [itex] f^{-1}(f(x))=x [/itex] for all [itex] x \in U [/itex].

    In conclusion, if [itex] f(x)=y [/itex], then so long as [itex]y[/itex] is in [itex] f(U) [/itex], we have [itex] x=f^{-1}(y) [/itex].

    In short... you can always solve [itex] y=f(x) [/itex] for [itex] x [/itex] as a function of [itex] y [/itex] somewhere, as long as [itex] f [/itex] is continuously differentiable and isn't a constant function.
     
    Last edited: Jun 28, 2012
  10. Jun 28, 2012 #9
    Well.. A neat lil trick you can do is (given f^-1=g), f(g(x))=x, derivative of bother sides is f'(g(x))*g'(x)=1, proceed from there :)
    (assuming invertibility)
     
    Last edited: Jun 28, 2012
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