## Calculus with Applications (10th Edition)

$\frac{1}{5}$
$\frac{x^{2}+y}{x-y}=9$ $x^{2}+y=9x-9y$ $2x\frac{dx}{dt}+\frac{dy}{dt}=9\frac{dx}{dt}-9\frac{dy}{dt}$ $(9-2x)\frac{dx}{dt}=10\frac{dy}{dt}$ Now substitute x =4, y = 2, and dx/dt =2 to get $\frac{dy}{dt}=\frac{1}{5}$