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Homework Help: Question about the use of Leibniz notations …

  1. Oct 25, 2007 #1
  2. jcsd
  3. Oct 25, 2007 #2

    Gib Z

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    Please don't take offense, and by the way welcome to Physics forums, but i very strongly recommend that you borrow a nice calculus book from the library, or purchase one if you are able to, because I can see from your other post that you have no previous experience with the topic. The parts of calculus your posts are on are quite far apart in level of difficulty, and you should really start with the basics. mathwonk recommends "calculus made easy" often for beginners, though I can't recall the author, try googling it.
  4. Oct 25, 2007 #3
    I have math books, but this i could just not understand this ...
    :(, well i know that i suck at math ..

    If i have understand it correct, can i write "u(x)*dy/dx" as:

    Kindly Pellefant ... if that is so then i will be pleased ...

    Anyway i think you may ahve a point in what you said ...
    Last edited: Oct 25, 2007
  5. Oct 25, 2007 #4
    Also i have read math in university level but it was awhile ago now, and then i just studied before the exame. Now i wont to put my math knowledge back on track, but you must laugh at me and thinking i am a idiot ...
  6. Oct 25, 2007 #5
    The article states that

    u*dy/dx + y*du/dx = d(y*u)/dx

    its the product rule
  7. Oct 25, 2007 #6


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    And "u(x)*y(x)d/dx" makes no more sense than "u sin(y)= uy sin"!

    "d/dx" is an operator- it has to be applied to something- and that something, by standard notation is written on the right.
  8. Oct 27, 2007 #7
    I see the reason i got that idea was because my math book wrote dy/dx as (d/dx)*y ...

    So my math book stated dy/dx = (d/dx)*y
  9. Oct 27, 2007 #8
    Double check what the text book actually wrote if it wrote (d/dx)*y to mean dy/dx then throw it away.

    More likely you have misinterpreted the text. Is it possible to scan the page and include it in your next post?
  10. Oct 27, 2007 #9
    kk but u can write it as:

    du/dx= d/dx (x^-1) where u = (x^-1)

    And anouther question, i wonder purely academically, would it be ok to do the following




    Oki i think i get much of it but (if you find me annoying just ignore to reply :))

    I don't get how he can make this assumption
    u(x) dy / dx + u(x) P(x) y = y du / dx + u dy / dx
    which would mean
    du / dx = u(x) P(x)

    My question is if this proof from the first link in the topic is complete? ...
    Last edited: Oct 27, 2007
  11. Oct 28, 2007 #10
    I wonder if many of those proof r correct purely mathematicaly ... here is anouther one ...

    http://www.bio.brandeis.edu/classes/biochem102/hndDiffEq.pdf [Broken]

    look at (3) before they can put out the constant of integration, they has to do the integration, right?
    Last edited by a moderator: May 3, 2017
  12. Oct 28, 2007 #11


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    1: I'll bet the book does NOT have that "*" which implies multiplication. What (d/dx)y means is "the differentiation operator applied to the function y".

    2: And did you notice that the y is to the right of the d/dx ?

    It may be you are confusing (d/dx)y with multiplication which is commutative. Applying an operator is not- as in my example above "sin x" is NOT the same as "x sin"!
  13. Oct 28, 2007 #12
    Sorry, my fault

    ~ Pellefant ...

    /and thank you for your reply, this has been learning for me ...
    Last edited: Oct 28, 2007
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