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What does find the dx/dy of (eq inside) for x>0 mean?

  1. Jun 7, 2013 #1
    1. The problem statement, all variables and given/known data

    find dy/dx for the function xy^2+xlnx=4y for x>o

    2. Relevant equations

    xy^2+xlnx=4y

    3. The attempt at a solution

    I found the dy/dx to be "(y^2+lnx+1)/(4-x)=dx/dy"
    What is it asking me to do with the x>0? Do they want me to plug in a value greater than 0 for x and solve for y? or what?
     
    Last edited: Jun 7, 2013
  2. jcsd
  3. Jun 7, 2013 #2
    The function ln, as you might know, isn't defined for values less than zero. Hence it is meaningless to speak of a function value, let alone a derivative there.

    By the way, for this to make sense to me, you have to have an equal sign somewhere. Like xy^2+xlnx = 0 or something - else it's just a value, not anything to define a function y(x).
     
  4. Jun 7, 2013 #3
    Sorry, the eq is actually xy^2+xlnx=4y. Thank you for pointing out the mistake, I'll edit it out.
     
  5. Jun 7, 2013 #4
    I get a slightly different answer. Check your derivative of the first term on the left.
     
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