For the circle $x^2 + y^2 = 9$, what is $\frac{d^2y}{dx^2}$ in terms of $y$?

Explanation

1. 01

   $2x + 2y\frac{dy}{dx} = 0$, so $\frac{dy}{dx} = -\frac{x}{y}$.

2. 02

   By the quotient rule, $\frac{d^2y}{dx^2} = -\frac{y - x\frac{dy}{dx}}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2} = -\frac{x^2 + y^2}{y^3}$.

3. 03

   With $x^2 + y^2 = 9$, $\frac{d^2y}{dx^2} = -\frac{9}{y^3}$.