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Thread: Implicit differenriarion

  1. #1
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    Implicit differenriarion

    Hello All,

    I'm working on differentiations right now and I'm having a little trouble trying to fully understand. I did this one, but I'm not sure if it's right.

    <br />
x^2y=x+y^2

    \frac{d}{dx}[x^2y]=\frac{d}{dx}[x]+\frac{d}{dx}[y^2]

    2xy\frac{dy}{dx}[y]=x+\frac{dy}{dx}[2y]

    <br />
\frac{dy}{dx}(y-2y)=x-2xy

    <br />
\frac{dy}{dx}=\frac{x-2xy}{y-2y}

    Did I do it right?


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  2. #2
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    Quote Originally Posted by drkidd22 View Post
    Hello All,

    I'm working on differentiations right now and I'm having a little trouble trying to fully understand. I did this one, but I'm not sure if it's right.

    <br />
x^2y=x+y^2

    \frac{d}{dx}[x^2y]=\frac{d}{dx}[x]+\frac{d}{dx}[y^2] Good

    2xy\frac{dy}{dx}[y]=x+\frac{dy}{dx}[2y]" alt="2xy\frac{dy}{dx}[y]=x+\frac{dy}{dx}[2y]" /> Product rule on left hand side... and, what is the derivative of x with respect to x?

    <br />
\frac{dy}{dx}(y-2y)=x-2xy

    <br />
\frac{dy}{dx}=\frac{x-2xy}{y-2y}

    Did I do it right?

    Sorry I can't get the red to work, will post more in a tic

    Edit:

    Just in case a picture helps...



    ... where



    ... is the product rule, and



    the chain rule. (Straight continuous lines differentiate downwards with respect to x, the straight dashed line similarly but with respect to the dashed balloon expression which is the inner function for the chain rule.)


    _________________________________
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    Last edited by tom@ballooncalculus; Aug 25th 2009 at 01:27 PM.
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  3. #3
    MHF Contributor Calculus26's Avatar
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    The LHS is incorrect--use product rule:

    d(x^2y)/dx = 2xy + x^2dy/dx

    The Rhs is 1 +2ydy/dx


    2xy + x^2dy/dx = 1 +2ydy/dx


    now finish
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  4. #4
    MHF Contributor Calculus26's Avatar
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    In general for implicit differentiation

    1. differentiate fns of x normally d(x^2)/dx = 2x

    2. when differentiating fns of y diff with respect to y and multiply by dy/x

    d(y^2)/dx = 2y dy/dx
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  5. #5
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    \frac{dy}{dx}=\frac{1-2xy}{x^2-2y}
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  6. #6
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    Yep
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  7. #7
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    thanks, I liked the diagram you posted. Look my other thread coming up. I got another one.
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