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Homework Help: Differentiation Help

  1. Jun 8, 2006 #1
    There are two differentiation questions that I am working on and was wondering if I did it right and if I did it wrong, to get some help:

    1) y=(2x^2-x)(3x^2+5)^8

    let u = (2x^2-x) and let v=(3x^2+5)^8

    y'=uv' + u'v

    u'=4x-1
    v'=48x(3x^2+5)^7

    y'=(2x^2-x)(48x)(3x^2+5)^7+(4x-1)(3x^2+5)^8


    2) This one was the one I was stuck on. I tried using quotient and product rule:

    y=(t^2-6t/t^2+6t)^5

    y'=5(t^2-6t/t^2+6t)^4 . 1[(2t-6)(t^2+6t)-(t^2-6)(2t+6)/(t^2+6t)^2

    Sorry if it's a bit messy. I still haven't gotten the hang of using LaTex yet >_<
     
  2. jcsd
  3. Jun 8, 2006 #2

    Hootenanny

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    First one is spot on.

    Is this what the second one is meant to look like?

    [tex]y= \left( \frac{t^{2} - 6t}{t^{2} + 6t} \right)^{5}[/tex]
     
  4. Jun 8, 2006 #3
    For the 2nd one, yessir!
     
  5. Jun 8, 2006 #4

    Hootenanny

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    You can simplify it a bit first;

    [tex]y= \left( \frac{t^{2} - 6t}{t^{2} + 6t} \right)^{5} = \frac{(t-6)^5}{(t+6)^5}[/tex]
     
  6. Jun 8, 2006 #5
    so should I just use power rule on the top and bottom? or use both quotient and product rule?
     
  7. Jun 8, 2006 #6

    Hootenanny

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    I would use the quotient rule on this one.
     
  8. Jun 9, 2006 #7

    HallsofIvy

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    Or write it as
    [tex](t-6)^5(t+6)^{-5}[/tex]
    and use the product rule.

    You would not, of course, just differentiate numerator and denominator separately!
     
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