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Why is there a dx/dx in the Implicit Differentiation rule?

  1. Mar 1, 2012 #1
    I tried deriving this one on my own and I'm just not understanding where the dx/dx term comes from. I'm looking dy/dx.

    Starting with F(x,y) = 0:

    [itex]\frac{\partial{F}}{\partial{x}}\frac{dx}{dx} + \frac{\partial{F}}{\partial{y}}\frac{dy}{dx} = 0[/itex]

    It seems redundant to say dx/dx when it turns out to be one anyway. Why does this step need to be done? Thanks
  2. jcsd
  3. Mar 1, 2012 #2


    Staff: Mentor

  4. Mar 1, 2012 #3


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    It doesn't need to be done. What text did you see that in?
  5. Mar 1, 2012 #4
    Briggs Calculus. It just seemed kinda strange to say that x is a function of x so we need to take the derivative of it.
  6. Mar 1, 2012 #5


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    i do it! o:)
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