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Math Help - What is the second derivative of xy+y^3=1 using implicit differentiation?

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    What is the second derivative of xy+y^3=1 using implicit differentiation?

    y'' of xy+y^3=1
    Thanks!
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    Re: What is the second derivative of xy+y^3=1 using implicit differentiation?

    Quote Originally Posted by citcat View Post
    y'' of xy+y^3=1
    what did you get for y' ?
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    Re: What is the second derivative of xy+y^3=1 using implicit differentiation?

    -(y/x-y)
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    Re: What is the second derivative of xy+y^3=1 using implicit differentiation?

    Hello, citcat!

    You're off to a bad start . . .

    Exactly where is your difficulty?
    Do you have the correct answer?


    \text{Find }y''\text{ for: }\:xy+y^3\:=\:1

    We have: . 1\!\cdot\!y + x\!\cdot\!y' + 3y^2\!\cdot\!y' \:=\:0

    . . . . . . . . . . . . x\!\cdot\!y' + 3y^2\!\cdot\!y' \:=\:-y

    . . . . . . . . . . . . . (x + 3y^2)y' \:=\:-y

    . . . . . . . . . . . . . . . . . . . y' \:=\:\frac{\text{-}y}{x+3y^2} .[1]

    Then: . y" \;=\;\frac{(x+3y^2)(\text{-}y') - (\text{-}y)(1 + 6yy')}{(x+3y^2)^2}

    . . . . . y'' \;=\;\frac{-xy' - 3y^2y' + y + 6y^2y'}{(x+3y^2)^2}

    . . . . . y'' \;=\;\frac{y - xy + 3y^2y'}{(x+3y^2)^2}


    Substitute [1]:
    . . y'' \;=\;\frac{y - xy + 3y^2\left(\frac{\text{-}y}{x+3y^2}\right)}{(x+3y^2)^2}

    . . y'' \;=\;\frac{y(x+3y^2) - xy(x+3y^2) - 3y^3}{(x+3y^2)^3}

    . . y'' \;=\;\frac{xy + 3y^3 - x^2y - 3xy^3 - 3y^3}{(x+3y^2)^3}

    . . y'' \;=\;\frac{xy - x^2y - 3xy^3}{(x+3y^2)^3}
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