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Find properties of rates of change of x^3 + y^3 = 1

Given the equation

    \[ x^3 + y^3 = 1. \]

This defines y as a function of x.

  1. Without solving for y show that the derivative y' satisfies the equation:

        \[ x^2 + y^2 y' = 0. \]

    (Assume the derivative y' exists.)

  2. Show that

        \[ y'' = -2xy^{-5} \]

    whenever y \neq 0. (Assume the second derivative y'' exists.)


  1. For this part we differentiate each side with respect to x (keeping in mind that y is a function of x, so we need to use the chain rule to differentiate y^3).

        \begin{align*}  x^3 + y^3 = 1 && \implies && 3x^2 + 3y^2 y'&= 0 \\  && \implies && x^2 + y^2 y' &= 0. \end{align*}

  2. Using part (a) we differentiate y' to find y'',

        \begin{align*}  &&y' &= -\frac{x^2}{y^2}\\ \implies && y'' &= \frac{-2xy^2 + x^2 2y y'}{y^4} \\  &&&= \frac{-2xy^2 + x^2 2 y \left( -\frac{x^2}{y^2} \right)}{y^4} \\  &&&= \frac{-2xy^3 - 2x^4}{y^5} \\  &&&= \frac{-2x(y^3 + x^3)}{y^5} \\  &&&= -2xy^{-5}  &(\text{since } x^3 + y^3 = 1). \end{align*}

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