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 $x$ .

Without solving for $y$ show that the derivative $y'$ satisfies the equation:
$x^2 + y^2 y' = 0.$

(Assume the derivative $y'$ exists.)
Show that
$y'' = -2xy^{-5}$

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

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