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Differentials - I'm HOPELESSLY Confused. PLEASE Help. PLEASE

  1. Oct 3, 2005 #1
    Please for the love of god help me.

    I have a fundamental misunderstanding of differentials and Leibniz notation. I'm confused as to even where I should begin. Please allow me to start off my explanation by showing how my book introduces u-substition:

    I have highlighted in red the parts I am confused on. Let me start... Why are we dealing with differentials!? Aren't differentials supposed to measure the "change in linearization"?! Differentials were taught to me in the context of approximations, for the life of me I cannot see the connection with integrals. Where do differentials and approximations and errors come into play here? Why am I dealing with differentials? On top of all of this I still do not understand why dy/dx = f'(x) can be written as dy = f'(x)dx (since I'm told dy/dx is not actually a fraction), which is just adding to my confusion. I am told that Leibniz was wrong on some things and that we do not take the dx stuff seriously and that I should ignore it, but it's right in this context??? Or it's sort of right in a way but not really?

    And how can you just interpret it AS IF it were a differential? What does that actually mean? Interpret "as if"?!?! This doesn't sound like math, it sounds like fancy handwaving that somehow produces a correct answer through magic. How do they just "interpret" something as something else? I am so lost that I'm about ready to fall out of my chair.

    Please somebody help. Spare no detail... I have searched Google until the early hours of the morning for weeks like a madman trying to figure this out.
  2. jcsd
  3. Oct 3, 2005 #2
    I think they are just assuming that for small values of h, the derivative becomes:

    [tex]f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\stackrel{\text{small h}}{\approx} \frac{f(x+h)-f(x)}{h}[/tex]

    In this case you can think of h as dx and f(x+h)-f(x) as dy. Separate using algebra and you have your result.

    Maybe someone else can explain it a little better :smile:
  4. Oct 3, 2005 #3


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    To be honest, my advice is to simply treat it as a heuristic that helps you get right answers. Once you've used this device to figure something out, you can always go back and work through the problem rigorously using methods you do know.

    But, if you're really curious, there are things called differential forms. In some sense, a differential form is just something in front of which you can stick an integral sign (with a region of integration attached) to get a number. Then, from this "definition" you can work out their properties, and justify manipulations like you see in that problem.
  5. Oct 4, 2005 #4


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    dy/dx is just a fraction relating the change in y to the change in x, which yields slope. You can move them around as much as you want; they are values. It is just a way of writing f '(x) when you may be interested in examing the change in x and change in y seperately.
  6. Oct 4, 2005 #5
    I hope this helps

    Quantities such as dx,dy etc are called differentials.
    They signify a small(infinitesimal) change in corresponding quantities x,y

    It is true that dy/dx is not exactly a ratio but often we treat them as ratios
    for example when solving differential equations.
    A formal proof on why we can treat them as ratio requires complex operator algebra.

    It is better to treat d/dx as a operator.
    Similarly d can be treated as operator

    So d of some quantity is a infinitesimal change in it and we are justified in writing dy=f'(x)dx

    Often you come across such expressions



    Instead of using d/dx

    we could simply write 2udu+2dx=0 or udu+dx=0
    relating differentials.
    Notice how d was used as operator.
  7. Oct 4, 2005 #6


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    I would recommend you go back and reread the section on differentials in your calculus book. Certainly differentials can be used as an approximation to "small changes" but that is not their definition.
  8. Oct 4, 2005 #7


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    Well, the problem is that differentials aren't usually introduced in elementary calculus books (except as this heuristic for doing integrals).

    dy/dx is not a fraction, even though much of the notation used encourages that interpretation. There's an example matt grime likes to give, that looks something like:

    \frac{dx}{dy} \frac{dy}{dz} \frac{dz}{dx} = -1
  9. Oct 4, 2005 #8
    Thank you

    yes sure you are right.
    Thanks for pointing that out.

    But then when I am forming differential equations for a physical model why do I always use these quantities as small changes

    say small area between spheres dA=4*pi*r^2*dr
  10. Oct 4, 2005 #9
    I did read that section over...about ten times. :frown: I've scanned the section of my book (Calculus: Early Transcendentals, by James Stewart) so you can see exactly what they say: http://img24.imageshack.us/img24/8485/differentials1fp.png. And here is the first explanation of the dy/dx notation in my book: http://img24.imageshack.us/img24/7254/dydx0tx.png.

    A little while back my teacher was talking about Leibniz notation in integrals, and was saying that inevitably we will all come to ask one specific question: Why is that little dx thing there? Why can't we do without it? It makes sense in definite integrals (like the sum of areas of rectangles with height f(x) and width dx) but why is it that we must write integral(f(x))dx instead of integral(f(x)), and write u = g(x) and calculate a corresponding du = g'(x)dx rather than just du = g'(x). Then he took more time out to try and explain the arc length function and all this ds/dx stuff because we weren't really *understanding* why it worked in some cases (surface areas), just that it works. But he didn't actually explain why but rather said that in Math 280 (calculus 3) these things will be explained. :(
  11. Oct 4, 2005 #10
    why is that so? [the example you provided]?

    and is it true for all situations?
  12. Oct 5, 2005 #11
    "but why is it that we must write integral(f(x))dx instead of integral(f(x)), and write u = g(x) and calculate a corresponding du = g'(x)dx rather than just du = g'(x)". :([/QUOTE]

    Why? Because we can prove that (taking x = g(u) )
    [tex]\int_{g(c)}^{g(d)}f(x)dx = \int_c^df(g(u))g'(u)du[/tex].

    We can not prove that [tex]\int_{g(c)}^{g(d)}f(x)dx = \int_c^df(g(u))du[/tex].
  13. Oct 16, 2005 #12
    Riemann sums

    I'm a little late posting to this thread but I'd recommend to samh that you break it down to Riemann sums. Sure there are differential forms and such running around but it can be explained a bit simpler without them. Bear with me, I'm no good at latex-y stuff yet.
    Say you've got an integral

    I(f(x))dx over [a,b] (so I'm using I as my integral).

    Now you'd like to so some substitution thingy. Well move back to the sum formulation first:

    Ef(c_i)(x_i - x_(i-1)) where i runs from 1 to n (Here E is my capital sigma).

    So we've partitioned our interval [a,b]. But perhaps [a,b] is just the image of a differentiable function g(s):[c,d]->[a,b]. Then for every c_i in an interval [x_(i-1), x_i] there is a r_i in an interval [s_(i-1), s_i] so we can rewrite our sum as

    Ef(g(r_i))(g(s_i) - g(s_(i-1))),

    but g(s_i) - g(s_(i-1)) is approximately equal to g'(r_i)(s_i - s_(i-1)) because g is differentiable. Now our sum looks like

    Ef(g(r_i))g'(r_i)(s_i - s_(i-1))

    which upon the appropriate limit goes to the integral


    That's the most straightforward way of understanding the change of variables/u-substitution that I can think of. As long as we don't mind the delta-x going to a dx in the limit from the sum to the integral, then we can't mind the delta-u (or in my post, the delta-s) going to the du (ds).

    Now of course you're not always subbing u for x, but the argument works for more complicated substitutions (like the one samh gave).
  14. Oct 22, 2005 #13
    Thanks for the help guys. Sorry about the late reply but I've been really busy with school. I'm understanding things a lot more now. I've read your input, some more websites, and found this entry in the Dr.Math archives which helped a lot too. So things are better.

    I have another question about derivative notation. If [tex]f'(x)[/tex] is the first derivative, [tex]f''(x)[/tex] the second, [tex]f^(5)(x)[/tex] the fifth, what would [tex]f^(1/2)(x)[/tex] be?

    I can kind of think of it like you think of regular exponents to numbers. Like x^2 is x multiplied by itself and x^(1/2) is the number that equals x when multiplied by itself. Is this how you would interpret my previous question? Could [tex]f^(1/2)(x)[/tex] be the function whose derivative equals x, or in other words, the integral of f(x)? Then would [tex]f^(3/2)(x)[/tex] mean the integral of f's third derivative? But then wouldn't that be the second derivative...ahh confusing.
    Last edited: Oct 22, 2005
  15. Oct 23, 2005 #14
    The topic of interest is called "Fractional Calculus" and there are several articles in "Mathematics Magazine" an undergraduate math publication produced by the MAA. I've listed three below. I've read the third one by
    Thomas Osler and Marcia Kleinz and found it sufficient to quench my own thirst. Good luck.

    Mathematics Magazine: Volume 50, Number 3, Pages: 115-122 1977
    Fractional Calculus
    Bertram Ross

    Mathematics Magazine: Volume 68, Number 3, Pages: 183-192 1995
    Derivatives of Noninteger Order
    Kenneth S. Miller

    College Math Journal: Volume 31, Number 2, Pages: 82-88 2000
    A Child's Garden of Fractional Derivatives
    Thomas Osler and Marcia Kleinz
  16. Oct 24, 2005 #15
    Thanks a lot homology. I saw your message early yesterday but couldn't get to reading it until around 8pm. Anyway I stayed up until about 3am reading every single website I could find trying to make sense of it. Well after seven hours and about four pieces of paper I got it. I forgot to mention that this was an extra credit assignment. :cool: I was able to explain fractional derivatives and integrals to my teacher and got points for it! He asks us questions he doesn't think we can answer and gives extra credit points for them. So yes thankyou very much. :cool:
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