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Math Help - Using implicit differentation for second deriviate

  1. #1
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    Using implicit differentation for second deriviate

    Hello,

    I am stuck on the following problem:

    sorry i don't know how to write in math on here

    i have

    on the curve at (1, 1 ) using implicit differentiation for the eq
    rootx + rooty + y = 3

    the first question is find y' (1). I got the answer of -1/3 which is right. Now it asks for y''(1) and I can't figure it out for the life of me. The answer is 5/27 and I just can't get it.

    Thanks in advance.
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  2. #2
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    The first derivative is

    |\frac{1}{2\sqrt{x} = \frac{1}{2\sqrt{y}\times y' = 0

    Thre second derivative is

    \frac{-1}{2x^3/2} + y'(\frac{-1}{2y^3/2}\timey' + ( 1 + \frac{1}{sqrt{y} = 0

    Now substitute the values and find the result.
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  3. #3
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    The first derivative is
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  4. #4
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    i can't see what you put down it says error

    thanks for the response though!
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  5. #5
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    The first derivative is

    \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}\times y' = 0

    The second derivative is

    \frac{-1}{2x^3/2} + y'(\frac{-1}{2y^3/2}) + ( 1 + \frac{1}{sqrt{y})y" = 0

    Now substitute the values and find the result.
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  6. #6
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    whaaaaat. i can't really understand what you wrote
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  7. #7
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    Quote Originally Posted by Zippyz View Post
    whaaaaat. i can't really understand what you wrote
    The first derivative is

    1/(2x^1/2) + y'[1/(2y^1/2) + 1] = 0

    The second derivative is

    -1/2(2x^3/2) + y'[-1/2(2y^3/2)*y'] + y"( 1 + 1/2y^1/2) = 0
    Last edited by sa-ri-ga-ma; May 17th 2010 at 08:29 PM.
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  8. #8
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    EDIT:

     \frac{1}{2 \sqrt{x}} + \frac{1}{2 \sqrt{y}} \frac{dy}{dx} + \frac{dy}{dx} = 0

     \frac{dy}{dx} = \frac{-\sqrt{y}}{\sqrt{x} + 2 \sqrt{x}\sqrt{y}}

     \frac{d^{2}y}{dx^{2}} = \frac{-\frac{1}{2\sqrt{y}} \frac{dy}{dx} (\sqrt{x} + 2 \sqrt{x}\sqrt{y}) + \sqrt{y} (\frac{1}{2\sqrt{x}}+\frac{1}{\sqrt{x}} \sqrt{y} + \sqrt{x}\frac{1}{\sqrt{y}} \frac{dy}{dx})} {(\sqrt{x} + 2 \sqrt{x}\sqrt{y})^{2}}

    so  \frac{d^{2}y}{dx^{2}} (1,1) = \frac{5}{27} using the fact that  \frac{dy}{dx} (1,1) = - \frac{1}{3}
    Last edited by Random Variable; May 17th 2010 at 08:38 PM.
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  9. #9
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    d/dx[1/(2y^1/2)*(dy/dx)] = (-1/(4y^3/2)(dy/dx)^2 + [1/(2y^1/2)]*d^2y/dx^2
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