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Finding the derivative, quotient rule with natural log function.

  1. Feb 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Find y' of
    y= 1-3ln(7x)/x^4

    2. Relevant equations



    3. The attempt at a solution
    I used the quotient rule and got:
    y'=x^4*d/dx(1-3ln(7x)-(1-3ln(7x)*d/dx(x^4)/(x^4)2

    which is: x^4*(0-3*1/7x*7)-(1-3ln(7x))*4x^3/x^8
    simplified to: 3x^4/x-1+3ln(7x)*4x^3
    3x^3-4x^3+12x^3ln(7x)/x^8
    take out the x^3 from the denominator and the numerator and I get the answer:

    y=-1+12ln(7x)/x^5
    The website I use to do my calculus homework says this is wrong, where is my error?
     
  2. jcsd
  3. Feb 19, 2012 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You dropped a minus sign. You should get -3x^3-4x^3. Not 3x^3-4x^3.
     
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