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Differentiating Products

  1. Jul 29, 2008 #1
    Hello, in coming up with a general form for finding the derivative of a product there is one step I do not understand.

    if we let u, v and y be functions of x then:

    dy/dx = u(dy/dx) + v(du/dx) + du (dv/dx)

    From there we take the lim as dx -> 0 of each term and the textbook I am reading says:

    "Now let dx -> 0. Then du -> 0"

    I simply do not understand why it follows that du -> 0 because dx -> 0.

    With this assumption however we are able to find the simplified general form:

    dy/dx = u(dv/dx) + v(du/dx)

    I applogize for not using the fancy math notation symbols. I have never used them before so I don't really know what half of them mean. I hope you guys can understand my question better than I can...
  2. jcsd
  3. Jul 29, 2008 #2


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    Hi Dumbledore! :smile:

    (have a curly-wurly delta: δ)
    You mean dy/dx = u(dv/dx) + v(du/dx) + du (dv/dx).

    I think it means that if you move from x to x + δx, so that u v and y move to u + δu, v + δv, and y + δy,

    then obviously (y + δy) - y = (u + δu)(v + δv) - uv = uδv + vδu + δuδv.

    Now divide by δx: (uδv + vδu + δuδv)/δx

    and let δx tend to zero (so all the other δs tend to zero also), and you get:

    u dv/dx + v du/dx + δu dv/dx,

    = u dv/dx + v du/dx + 0.

    (or, of course, you can write it u dv/dx + v du/dx + δv du/dx.)

    That's the δu which they're talking about. :smile:
  4. Jul 29, 2008 #3
    I think the part that I was missing is that if δx approaches 0 then all the other δs approach zero. Now it makes perfect sense. Thanks tiny tim!
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