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Homework Help: The chain rule?

  1. Oct 1, 2006 #1

    [tex]u(x) = u_1(x)u_2(x)\cdot\cdot\cdot u_n(x)[/tex]

    Show that

    [tex]u'(x) = u(x)\left[\frac{u'_1(x)}{u_1(x)} + \frac{u'_2(x)}{u_2(x)} + ... + \frac{u'_n(x)}{u_n(x)}\right][/tex]

    Hint: Take the logarithm of u(x) first.

    I solved this by using the chain rule. Regarding the hint, am I missing something?

    Next problem states that, by using the formula for u'(x), differentiate

    [tex]u(r) = \frac{1}{(r+1)(r+2)\cdot\cdot\cdot(r+n)}[/tex]

    When using the formula, my answer to this is

    [tex]u'(r) = -\frac{1}{(r+1)(r+2)\cdot\cdot\cdot(r+n)} \left[\frac{1}{r+1} + \frac{1}{r+2} + ... + \frac{1}{r+n}\right][/tex]

    but the correct answer should be

    [tex]u'(r) = (r+1)(r+2)\cdot\cdot\cdot(r+n)\left[\frac{1}{r+1} + \frac{1}{r+2} + ... + \frac{1}{r+n}\right][/tex]

    This seems like an elementary problem but I'm not getting the correct result so I must be missing something..?
  2. jcsd
  3. Oct 1, 2006 #2


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    Do you mean you didn't use the hint? The idea is to take the logarithm, then differentiate using the chain rule.

    This cannot be the correct answer.
    You can see this by looking at the problem from a physical point of view: the units don't match. (say r is in meters, then the units of u' should be 1/m^(n+1))
    Or from a mathematical point of view: it has only removable singularities, while u clearly goes to infinity at r=-i, i=1,2,..,n.

    Your answer is in fact correct.
  4. Oct 1, 2006 #3
    In the first part I didn't use the hint, I used the product rule.
    I'm not as stupid as I think. The book is wrong and I am right. Thanks. =)
  5. Oct 1, 2006 #4

    matt grime

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    What level of rigour do you require? If using the product rule, then I hope you used induction on the number of terms, say, unless you;re allowed to use the product rule for n factors (which is entirely possible). Using logs it is a trivial exercise since logs turn products into sums and differentiation is linear.
  6. Oct 1, 2006 #5
    Am I not allowed to do it like this:

    [tex]u'(x) = u'_1 u_2 \cdot \cdot \cdot u_n + u_1 u'_2 \cdot \cdot \cdot u_n + ... + u_1 u_2 \cdot \cdot \cdot u'_n = u \left[ \frac{u'_1}{u_1} + \frac{u'_2}{u_2} + ... + \frac{u'_n}{u_n} \right][/tex]

    Last edited: Oct 1, 2006
  7. Oct 1, 2006 #6


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    Yes, you are allowed to do that. That is where induction with multiple terms in the product rule would ultimately take you as to which matt alluded.
  8. Oct 1, 2006 #7
    That's funny, that is differentiating method that Feynman liked to use.
  9. Oct 1, 2006 #8
    Yes that's funny. Maybe I have some potential. ;)
  10. Oct 1, 2006 #9


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    The product rule? I imagine a log of people like to use that method!

    If [itex]u(x)= u_1(x)u_2(x)\cdot\cdot\cdot u_n(x)[/itex] then
    [itex]ln(u(x))= ln(u_1(x))+ ln(u_2(x)+\cot\dot\cdot + ln(u_n(x))[/itex]
    so that
    [tex]\frac{1}{u(x)}u'(x)= \frac{1}{u_1}u_1'(x)+ \frac{1}{u_2}u_2'(x)+ \cdot\cdot\cdot+ \frac{1}{u_n}u_n'(x)[/itex]
    Multiplying both sides of that by u(x) gives the result.
  11. Oct 1, 2006 #10


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    A LOG of people??
    Whatever base are you using here??:confused: :confused:
  12. Oct 2, 2006 #11
    Well, it is not entirely just the product rule. There are a few steps to get there, not that it is out of people's reach, it's just not as obvious as you are making it out to be that you can differentiate like that.

    The people log base, haha.
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