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Math Help - Urgent second differential problem

  1. #1
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    Urgent second differential problem

    Please someone please find the second differential for this equation

    (x3) + (y3) = 1
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    Re: Urgent second differential problem

    Quote Originally Posted by righteous818 View Post
    Please someone please find the second differential for this equation

    (x3) + (y3) = 1
    Try using implicit differentiation.
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    Re: Urgent second differential problem

    i did but i am not the gettin the answer the book said the answer is -2x/(y5)
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    Re: Urgent second differential problem

    Quote Originally Posted by righteous818 View Post
    i did but i am not the gettin the answer the book said the answer is -2x/(y5)
    In that case, please post all your working and we will help you get back on track.
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    Re: Urgent second differential problem

    how do i get fractions
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    Re: Urgent second differential problem

    If you are unsure how to use LaTeX, just use a / for the vinculum and use brackets where they are necessary. Otherwise inside tex tags, use the command \frac{}{}, with the numerator in the first set of brackets and the denominator in the second.
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    Re: Urgent second differential problem

    well for te first differential i got

    3x2 + 3y2(dy/dx)=0

    dy/dx = (-x2 )/(y2)

    Then seoncd differntial using quotient rule

    [(y2)(-2x)-(-x2 )(2y(dy/dx))]/(y4)

    then finally the second differntial came down to for me

    [-2xy3 + 2x4 ]/y5


    Please help
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    Re: Urgent second differential problem

    Quote Originally Posted by righteous818 View Post
    well for te first differential i got

    3x2 + 3y2(dy/dx)=0

    dy/dx = (-x2 )/(y2)

    Then seoncd differntial using quotient rule

    [(y2)(-2x)-(-x2 )(2y(dy/dx))]/(y4)

    then finally the second differntial came down to for me

    [-2xy3 + 2x4 ]/y5


    Please help
    You are correct up to where you have used the Quotient Rule to find the second derivative. So you have

    \displaystyle \begin{align*} \frac{d^2y}{dx^2} &= \frac{-2xy^2 + 2x^2y\,\frac{dy}{dx}}{y^4} \\ &= \frac{-2xy + 2x^2\frac{dy}{dx}}{y^3} \\ &= \frac{-2xy + 2x^2\left(-\frac{x^2}{y^2}\right)}{y^3} \\ &= \frac{-2xy - \frac{2x^4}{y^2}}{y^3} \\ &= \frac{-2xy^3 - 2x^4}{y^5} \end{align*}

    Your final answer has a sign error but is otherwise correct.
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    Re: Urgent second differential problem

    thanks
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