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Integration involving ln

  1. Nov 17, 2009 #1
    1. The problem statement, all variables and given/known data
    Find dy/dx of y=1/xlnx


    2. Relevant equations
    I thought that I'd have to use the quotient rule:
    y'=vu'-uv'/v^2

    However the differential of 1 is 0, so is this the right rule to use?

    3. The attempt at a solution

    u=1
    u'=0

    v=xlnx
    v'=1/x

    y'=vu'-uv'/v^2
    y'=(xlnx*0)-(1*1/x)/(xlnx)^2
    y'=-1/x(xlnx)^2
    y'= -1/x^3lnx^2

    But this doesnt seem right to me, I'm not sure. Any help at all would be greatly appreciated. Many Thanks.
     
  2. jcsd
  3. Nov 17, 2009 #2
    Really? I thought 1/x was the derivative of lnx, not xlnx.
     
  4. Nov 17, 2009 #3
    Yes sorry just realised that..
    I redid it and got dy/dx=1-lnx/(x^2lnx^2), which sounds a more accurate answer and think I have done it right this time.

    Is is possible to just extend this question though:
    If I was to find when the gradient equals 0, I've got 0=1-lnx so 1=lnx, but how would I solve this?

    Thankyou
     
  5. Nov 17, 2009 #4
    e to the what equals 1?
     
  6. Nov 17, 2009 #5
    Does lne=1??
    But I dont know how that solves this, does that mean x=e?
     
  7. Nov 17, 2009 #6
    Yes, ln(e)=1.

    I kindve messed up my hint there and ended up inadvertantly giving answer instead. (e^0 is 1, but thats not relevent here.)

    Remember ln is just log base e, which 'undoes' powers, so you you're looking for the power of e that equals e in this case. (e^1=e^ln(x) => e^(lnx) is just x, =>e=x).
     
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