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Leibniz notation giving me a headache

  1. Sep 6, 2007 #1
    ----EDIT------
    In my original post I totally messed up my variables. Here I'll get straight to the point.

    I guess what I am really asking is how is [tex]\frac{dy/dx}{y-900}[/tex] equal to [tex]\frac{d}{dx}[/tex]ln|y-900|

    I know that the integral of [tex]\frac{1}{y-900}[/tex] is ln|y-900| but I don't understand how the notation works where one side has dy/dx and the other side just has d/dx
     
    Last edited: Sep 6, 2007
  2. jcsd
  3. Sep 6, 2007 #2
    Yes what they are doing seems very confusing. In particular, where did t come from? It would seem obvious that the solution to your ode would be just to multiply both sides by (x-900) and then integrate to give y as a quadratic function of x. Is it possible that there is a typo, or a mistake in how you read the question?
     
  4. Sep 6, 2007 #3
    I don't know about the first part but for the second part just remember that Integrating a differential just leaves what's being being differentiated:

    [tex]y = x^2[/tex]
    [tex]\int \frac{dy}{dx} dx = y = x^2[/tex]

    If the Leibniz notation is giving you problems I think you can just change it to the hm..American Style ? my teacher calls it that...

    so you'd get:

    [tex]\frac{y'}{x-900} = \frac{1}{2}[/tex]

    [tex]y=ln|x-900|[/tex]

    [tex]y'=(ln|x-900|)'=\frac{1}{x-900}[/tex]

    which doesn't seem to be the same since the top y' on the first formula seems to disappear going to the second one... But then again I have no knowledge of Diffy Eq's.

    Also did they integrate w.r.t. x on the LHS but w.r.t. t on the RHS? on the final step of y'=1/2
     
  5. Sep 6, 2007 #4

    mathwonk

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    do you know the chain rule?

    it impies that d/dx (ln(f)) = (1/f)df/dx.
     
    Last edited: Sep 6, 2007
  6. Sep 6, 2007 #5
    I know the chain rule and I know how to do the computations, what I do not understand is what exactly the dy/dx and then later d/dx means.

    My problem was when I went through my calculus classes I always used f'(x) or y' for my notation. And then with integrals f__dx I just learned how to use it not what it actually means. It seems like there are so many different meanings for leibniz notation and no one really agrees on it. Today I asked a physics teacher how to solve a problem and he used dx and dy almost like variables and put one to one side and one to the other than slapped on an integral sign and that worked too...
     
  7. Sep 6, 2007 #6
    That's because you can do analysis/calculus in a variety of ways -- to be really fancy, "analysis is not fully identified in first order logics". Once you get to learn "real" analysis, you'll come across a metric buttload of things like epsilon and deltas and taking limits as things tend to zero, or this or that. There's another approach, ironically called "non-standard analysis", in which we extend the real numbers, and have infinitesimals that behave rather like what Leibniz notation suggests, except with actually self-consistent algebraic rules.

    Personally, I'm a physicist, so I just throw dx and dy about like they're variables, occasionally throwing in an integral sign to the mix to make things correct. It's all really just shorthand for the long, formal manipulations. Dodgy shorthand, to be sure.
     
  8. Sep 6, 2007 #7
    This sort of helps ease my mind since I am always wondering, is this ok? Can we treat this as a variable sometimes and sometimes not. It doesn't seem like there is some real rule to follow... unless I guess I went further into math and did that other stuff.

    So for now I guess I just have to learn when it is ok and when it is not and just accept it? I'm a chemistry major just minoring in math so I don't really need any more math than just the application portion of it, at least for now.



    But I still don't understand the difference between dy/dx and just d/dx
     
  9. Sep 6, 2007 #8
    Strictly speaking, dy/dx is a variable (taken as a whole, no splitting!) and d/dx is an operator. If you know anything about programming, operators are like higher-order functions -- they take functions and spit out another one.
     
  10. Sep 6, 2007 #9
    Y s just some differentiable function. It’s an arbitrary symbol used to represent a function. As long as you get that in your head you will be fine.

    dy/dx by definition is the derivative of y with respect to x.

    But what if y = x^2 + e^x

    Then couldn’t we write dy/dx = 2x + e^x or d(x^2 + e^x) = 2x + e^x?

    Putting d/dx is just short hand for the second way we write it to mean take the derivative of what ever function follows.
     
  11. Sep 6, 2007 #10
    Thanks for all the help guys, I'm pretty sure I get it now :)
     
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