## University Calculus: Early Transcendentals (3rd Edition)

$∂f/∂x = 2x(y+2)$ $∂f/∂y = x^2-1$
$f(x,y) = (x^2-1)(y+2)$ 1. Find $∂f/∂x$: Using the chain rule: $∂f/∂x = (x^2-1)'(y+2) + (x^2-1)(y+2)'$ **$∂/∂x(x^2-1) = 2x$ **$∂/∂x(y+2) = 0$ $∂f/∂x= 2x * (y +2) + (x^2-1)(0)$ $∂f/∂x = 2x(y+2)$ 2. Find $∂f/∂y$: Using the chain rule: $∂f/∂y = (x^2-1)'(y+2) + (x^2-1)(y+2)'$ ** $∂/∂y(x^2-1) = 0$ ** $∂f/∂y(y+2) = 1$ $∂f/∂y = 0(y+2) + (x^2-1)*(1)$ $∂f/∂y = x^2-1$