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How to take the derivative of implicit functions

  1. Feb 24, 2009 #1

    Mentallic

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    I have been able to follow how to take the derivative of implicit functions, such as:

    [tex]x^2+y^2-1=0[/tex]

    Differentiating with respect to x

    [tex]2x+2y\frac{dy}{dx}=0[/tex]

    [tex]\frac{dy}{dx}=\frac{-x}{y}[/tex]

    Sure it's simple to follow, but I don't understand why the [tex]\frac{dy}{dx}[/tex] is tacked onto the end of the differentiated variable y.

    An explanation or article on the subject would be appreciated. Thanks.
     
  2. jcsd
  3. Feb 24, 2009 #2

    cristo

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    You're differentiating wrt x, so using the chain rule:

    [tex]\frac{d}{dx}(y^2)=\frac{d}{dy}(y^2)\frac{dy}{dx}=2y\frac{dy}{dx}[/tex]
     
  4. Feb 24, 2009 #3

    Mentallic

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    Aha, so it's done using the chain rule. Thankyou :smile:
     
  5. Feb 24, 2009 #4

    HallsofIvy

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    It's not "tacked on", it's nailed firmly!:tongue2:
     
  6. Feb 25, 2009 #5

    Mentallic

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    haha :rofl:
    I always think 2 moves ahead, taking into consideration that separating to isolate will be necessary. Nail vs tack, I think we know the winner :wink:
     
    Last edited: Feb 25, 2009
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