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How to differentiate the expression 4xy = y^2 + 2ln(x)

  1. Mar 19, 2005 #1
    [tex]$ 4xy = y^2 + 2 \ln x[/tex]
    How do I differentiate that?
     
  2. jcsd
  3. Mar 19, 2005 #2
    Use implicit differentiation.

    4y + 4xy' = 2yy' + 2/x
    y'(4x - 2y) = 2/x - 4y
    y' = (2/x - 4y)/(4x - 2y)
     
  4. Mar 19, 2005 #3
    :confused: :confused: Sorry, but I've not learned implicit differentiation. Haven't even heard of it.
     
  5. Mar 19, 2005 #4
    Then I don't know how you would isolate y in that to solve it the normal way.
     
  6. Mar 19, 2005 #5

    t!m

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    Rewrite the expression as [tex]y^2 - 4xy + 2 \ln x = 0[/tex] and use the quadratic formula to find y in terms of x. Then derive.
     
  7. Mar 19, 2005 #6
    Got it. Thank you.
     
  8. Mar 19, 2005 #7
    Why do teachers ask their students to do something like that when it can be solved using a more proficient technique?
     
  9. Mar 19, 2005 #8

    Hurkyl

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    Staff Emeritus
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    Gold Member

    It's always reassuring to see that two drastically different methods yield the same results.

    Also, doing this problem both ways will allow students to greater appreciate implicit differentiation, once they see how much work it can save.
     
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