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Implicit diff.

  1. Sep 3, 2008 #1
    Heres another problem I was working on....

    http://img141.imageshack.us/img141/2318/calc2qg4.jpg

    Im trying to find dy/dx using implicit differentiation.....my algebra is a bit rusty......but Im trying to make sure im on the right track....
     
  2. jcsd
  3. Sep 3, 2008 #2
    Implicit diff of x^2 is not 2xy', it's just 2x. Everything else is fine.
     
  4. Sep 3, 2008 #3
    where did I do that in the problem?
     
  5. Sep 3, 2008 #4
    Not quite;

    When you are differentiating x2 with respect to x, for example, you simply get 2x. You do not multiply 2x by dy/dx.

    However, when you (implicitly) differentiate y2 with respect to x (where y is a function of x) you get

    [tex]
    2y \frac{dy}{dx}
    [/tex]

    Here you do multiply by dy/dx; the reason comes from the chain rule.
     
  6. Sep 3, 2008 #5
    Third line, first expression on the left hand side (referring to your attached image)
     
  7. Sep 3, 2008 #6
  8. Sep 3, 2008 #7
    does the work above look ok?
     
  9. Sep 3, 2008 #8
    Looks fine to me
     
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