## Calculus: Early Transcendentals (2nd Edition)

Slope at $(1,1) = -1$
$x^{\frac{2}{3}}+y^{\frac{2}{3}} = 2$ First Derivative: $\frac{2}{3}x^{\frac{-1}{3}} + \frac{2}{3}y^{\frac{-1}{3}}\frac{dy}{dx} = 0$ $\frac{dy}{dx} = \frac{-y^{\frac{1}{3}}}{x^{\frac{1}{3}}}$ Slope at $(1,1)$ $\frac{dy}{dx} = \frac{-1^{\frac{1}{3}}}{1^{\frac{1}{3}}} = -1$