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Math Help - implicit differentiation

  1. #1
    Newbie midsummer's Avatar
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    implicit differentiation

    I tried doing this myself but I'm really not sure my answer makes sense. :/

    find dy/dx by implicit differentiation
    square root (x+y)=1 +x^2 y^2
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by midsummer View Post
    I tried doing this myself but I'm really not sure my answer makes sense. :/

    find dy/dx by implicit differentiation
    square root (x+y)=1 +x^2 y^2
    Consider y as a function with respect to x.

    So you have \sqrt{x+y}
    What is its derivative ? It's \frac{(x+y)'}{2 \sqrt{x+y}} (using chain rule)
    What is (x+y)' ? It's (x)'+(y)'=1+y'
    So (\sqrt{x+y})'=\frac{1+y'}{2 \sqrt{x+y}}

    You have 1+x^2 y^2. What I advise you is to write 1+(xy)^2
    What is its derivative ? Using chain rule, we have : (1)'+2(xy)'(xy)=2(xy)'(xy)
    What is (xy)' ? Using product rule, we have : y+xy'
    So (1+(xy)^2)'=2(y+xy')(xy)


    \boxed{\frac{1+y'}{2 \sqrt{x+y}}=2(y+xy')(xy)}

    Solve for y'=\frac{dy}{dx}


    ******** chain rule says : f(g(x))=g'(x) f'(g(x))
    Last edited by Moo; October 9th 2008 at 11:49 AM.
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  3. #3
    Newbie midsummer's Avatar
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    Great explanation. You are awesome!
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