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Quotient rule for derivatives (algebraic division help?)

  1. Jun 24, 2004 #1
    Hi, I'm currently reading "Calculus Made Easy" and ran into a road block. I'm reading this in my spare time so it's not school work or anything (I know some forums have policies about this is why I mention). The answer is there. I just want to know how to go about doing this problem. It relates to the quotient rule for differentiating but I don't think that is terribly important since it's just some simple algebra that's eluding me.

    Just an explanation or a link with information on how to do this would be nice. I am familiar with algebraic division (not a ton, but some) and couldn't figure it out. Neither could my friend who is decent with algebra. Anyway enough boring you. Thanks in advance.

    edit: I know how to get the first equation, I just need to know how to do the division part if it wasn't clear.

    Attached Files:

  2. jcsd
  3. Jun 24, 2004 #2


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    How old is that book? Looks like a very "old style" analysis. Basically, what is happening is "long division": to divide u+ du by v+ dv, how many times does v divide into u: u/v times of course. Now multiply v+dv by u/v: u+ udv/v and subtract from u+ du: u+ du- (u+ udv/v)= du- udv/v. How many times does v divide into that? du/v.
    Multiply (du/v)(v+ dv)= du+ dudv/v, subtract that from du- udv/v and continue.

    This book is called "Calculus Made Easy"? Looks like their idea of easy is being very formal and ignoring what the symbols really mean. In particular, unless you are going to non-standard analysis (very hard!) it is a very bad idea to treat differentials (dx, dy, etc.) as if they were regular numbers- it works but the reason it works is very deep.

    If I were going to calculate the differential of y= u/v using that method (which I have just said is a bad idea) here is what I would do: y+ dy= (u+ du)/(v+ dv) (by definition). Subtract y from both sides: dy= (u+ du)/(v+ dv)- u/v= v(u+ du)/((v)(v+ dv)- u(v+dv)/(v(v+dv))= (uv+ vdu-uv-udv)/(v2+vdv)= (vdu- udv)/(v2- vdv). But since vdv is an "infinitesmal" compared with v2 we can ignore it in the sum v2- vdv (that's the part that makes this a "very bad idea"- proving that involves either using non-standard analysis or a limit process which this method is ignoring) so we have dy= (vdu-udv)/v2.
  4. Jun 24, 2004 #3
    Thanks for your help. Yes, the book is pretty old. I believe over 75 years old, revised a few times. The reason I bought it is that it received good reviews and people said it was a recommended for a decent intro to calculus. It's turning out to be a pain in areas though because of the older style of a lot of things. I've ended up using my knowledge from Morris Kline's Calculus to reason what he's saying in this one, sometimes (stuff like "just throwing things away" with no real reasoning). Too bad I can't locate an answer key, to check my work, for Morris Kline's book or I'd just stick with that. :frown:
  5. Jun 25, 2004 #4
    I originally bought that book based on the positive reviews it received. It doesn't deserve them. It sucks. By contrast to the other books available a hundred years ago, it was probably great, but it's extremely outdated and difficult to read now. The "preview" on Amazon.com was only of the earliest chapters, which were a recent addition by a more modern author and do not reflect the majority of the book's content.
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