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Evaluating derivative of an equation

  1. Nov 6, 2012 #1
    1. The problem statement, all variables and given/known data

    y = (x^2+1)^1/2 * (x^3+1)^1/3 * (x^4+1)^1/4 * ....... * (x^100+1)^1/100

    y'(1) Evaluate exactly.

    2. Relevant equations

    3. The attempt at a solution

    I'm not exactly sure what is needed to solve this, but I tried using product/chain rule but that doesn't end up nicely at all.

    I also tried using natural logs/implicit differentiation which was better but still couldn't finish it:

    y = 1/2ln(x^2+1) + 1/3ln(x^3+1)

    dy/dx = y [1/2(2x/x^2+1) + 1/2(3x^2/x^3+1)......]

    So basically dy/dx of all the terms on the right are 1/2 when x=1?
    If so, then
    dy/dx = y [99*1/2] = y[49.5]

    But it says to evaluate exactly so i'd need to plug in y which makes things complicated again? Any help would be appreciated, sorry if I didnt' make this very clear.
     
  2. jcsd
  3. Nov 6, 2012 #2

    SammyS

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    That should be ln(y) = 1/2ln(x^2+1) + 1/3ln(x^3+1) + ... + 1/100ln(x^100+1)
    That should be dy/dx = y [1/2(2x/(x^2+1)) + 1/3(3x^2/(x^3+1)) + ... + 1/100(100x^99/(x^100+1))]
    But your result is correct.

    What is y(1) ?
     
  4. Nov 6, 2012 #3
    Yea I guess it was lazy on my part.

    That's the problem, I can't figure out a way to find out y(1) easily..

    I know that it plugging 1 in would result in 2^(1/2) * 2^(1/3) * 2^(1/4).... *2^(1/100) and I could just add the exponents but that would take too much effort unless I'm forgetting something that I probably learned way back?
     
  5. Nov 6, 2012 #4

    SammyS

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    Isn't ak∙am∙an = ak+m+n ?

    ... so, what is 1/2 + 1/3 + 1/4 + 1/5 + ... + 1/100 ?

    I know. It's still pretty messy.
     
  6. Nov 7, 2012 #5

    Hmm that's what i meant but yea i forgot how to do that.. but ill try to figure it out. We get about a third of a page to show all our work so i was just thinking there was an easier way to solve this. thanks for your help!
     
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