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Implicit differentiation question: can't divide a fraction divided by another

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

    I'm try to implicitly differentiate the function: xlny+√y=lnx


    3. The attempt at a solution

    And I got to the stage where I have: dy/dx = (1/x-lny)/(x/y+1/(2*√y)) which is where I have no idea on how to clean this up. Could someone please explain to me how to simplify a function like that?

    Im not quite sure if I was heading to the right anwer but anyway the book provided the answer of
    (2y-2xylny)/(2x^2+x√y).

    Either way I would like to know how to simplify an equation like the one I have as I've come across it a few times.
    Thanks for your help in advance.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 1, 2012 #2

    Dick

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    Multiply numerator and denominator by 2xy. Do you see how this clears the denominators on the top and bottom? Can you see why 2xy is a good choice?
     
  4. Aug 1, 2012 #3
    I see so your chose 2xy to cancel everything, so when it comes to this situation you are allowed to choose any term that will simplify the equation? Of course as long as your multiply top and bottom.

    I'm still stuck now I'm up to dy/dx = [(2y-2xyln(y)]/(2x^2+2xy√y) its so close to the answer of (2y-2xylny)/(2x^2+x√y) how do I get rid of the 2 and the y from the last bit 2xy√y?
     
  5. Aug 1, 2012 #4

    Dick

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    2xy times 1/(2*sqrt(y)=x*sqrt(y), yes? And sure, you chose any factor that will simpify the result. Choose the minimal one.
     
  6. Aug 1, 2012 #5
    Wait disregard my last attempt I made a mistake. I now have [2y-2xln(y)]/[2x^2+(xy/√y)] and I need the last bit to look like (x√y).
     
  7. Aug 1, 2012 #6
    could I please ask why 2xy times 1/(2*sqrt(y)=x*sqrt(y)? I see the 2's cancel but what about the rest?
     
  8. Aug 2, 2012 #7

    Dick

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    y/sqrt(y)=sqrt(y). Because sqrt(y)*sqrt(y)=y. That's what sqrt means. If y is negative it doesn't really work, but that's not really the point here.
     
  9. Aug 2, 2012 #8
    Oh yea now I got you, thanks so much for your help.
     
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