## Calculus: Early Transcendentals (2nd Edition)

$\frac{d^2x}{dy^2} = -\frac{-1}{4y^3}$
$x+y^2=1$ Take Derivative and Solve for $\frac{dy}{dx}$ $1+2y\frac{dy}{dx}=0$ $\frac{dy}{dx} = \frac{-1}{2y}$ Take Second Derivative and Solve for $\frac{d^2y}{dx^2}$ $\frac{d^2x}{dy^2} = \frac{2}{4y^2}\times \frac{-1}{2y} = -\frac{-1}{4y^3}$