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

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


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

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


    ******** chain rule says : $\displaystyle f(g(x))=g'(x) f'(g(x))$
    Last edited by Moo; Oct 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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