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Homework Help: High School calculus problem, basic for some, need a little help

  1. Oct 13, 2008 #1
    Find the derivative of y=(3x+4)^5/(2x^2-3x)^6



    Some how the answer is simplified to 3x(6x^2-59x+24)/(2x-3)^7, I don’t know if I just copied it down wrong or I’m doing it incorrectly. Some assistance would be nice.

    I changed the equation to y=(3x+4)^5*(2x-3x)^-6 and used the product rule/ chain rule

    (3x+4)^5[-6(2x^2-3x)^-7(4x-3)]+(2x^2-3x)^-6[5(3x+4)^4(3)]

    This simplified into
    (-24x+18)(3x+4)^5divided by(2x^2-3x)^7 + 15(3x+4)^4 divided by (2x^2-3x)^6

    Multiplying 15(3x+4)^4 divided by (2x^2-3x)^6 by (2x^2-3x) to the bottom and top get a common denominator
    Which made it:
    [(-24x+18)(3x+4)^5]+[(3x+4)^4(30x^2-45x)] all over (2x^2-3x)^7

    At which point I took out the (3x+4)^4 which made it to:

    ((3x+4)^4) [(3x+4)(-24x+18)+(30x^2-45x] all over (2x^2-3x)^7

    After foiling and simplifying the inside of the brackets I got

    (3x+4)^4) (-72x^2-87x+72) taking out the three my final answer being:

    3(3x+4)^4 (-14x^2-29x+24) all over (2x^2-3x)^7

    I understand that in the answer they must have taken out an x at the bottom

    Sorry if my steps are a little confusing. Did I do it right or did I miss an important piece?
     
  2. jcsd
  3. Oct 13, 2008 #2
    the -14x^2 in my final answer should be -24, but yeah that does not change much.
     
  4. Oct 13, 2008 #3

    Dick

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    Your answer looks good to me. They took more than an x out of the bottom. (2x^2-3x)^7=x^7*(2x-3)^7. I think their answer is simply wrong.
     
  5. Oct 13, 2008 #4

    HallsofIvy

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    5 hours between your first and third posts? Do you expect people to be sitting around waiting for you to post a problem?

    Okay, that's a good idea. I have always preferred that to the quotient rule.

     
  6. Oct 13, 2008 #5
    No, I do not expect people to be waiting around for me to post a problem; I was merely annoyed that no one had replied in my five hours.

    Rather than the quotient rule is all I meant by my phrase.
     
  7. Oct 13, 2008 #6
    In my g, I forgot the square so it should be (2x^2-3x)^-6
    and consequently would make g’=-6(2x^2-3x)^-7(4x-3) correct?
     
  8. Oct 13, 2008 #7

    Dick

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    I wasn't checking all of the steps, but "3(3x+4)^4 (-14x^2-29x+24) all over (2x^2-3x)^7" is what I got.
     
  9. Oct 13, 2008 #8
    Ok, thank you very much, I must have written the answer down incorrectly.
     
  10. Oct 13, 2008 #9

    Dick

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    I'm quoting what you wrote down.
     
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