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Establish properties of y” given 3x^2 + 4y^2 = 12

Consider the equation

    \[ 3x^2 + 4y^2 = 12 \qquad \text{for } |x| \leq 2. \]

Show that y'' satisfies

    \[ 4y^3 y'' = -9. \]

(Assume the second derivative y'' exists.)


We differentiate (keeping in mind we must use the chain rule to differentiate y since it is a function of x),

    \begin{align*}  3x^2 + 4y^2 = 12 && \implies && 6x + 8yy' &= 0 \\  && \implies && y' &= -\frac{3x}{4y} \\  && \implies && y'' &= \frac{-12y + 12xy'}{16y^2} \\  &&&&&= \frac{-12y-\frac{9x^2}{4y}}{4y^2} \\  &&&&&= \frac{-12y^2 - 9x^2}{16y^3} \\  &&&&&= \frac{-3(4y^2 + 3x^2)}{16y^3} \\  &&&&&= -\frac{9}{4}y^{-3}. \end{align*}

Thus, 4y^3 y'' = 9.

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