## Calculus: Early Transcendentals 8th Edition

$y' = \frac{1 - e^y}{xe^y + 1}$
$xe^y = x - y$ use the product rule on $xe^y$ and differentiate both sides $(x)(y'e^y) + (1)(e^y) = 1 - y'$ $xy'e^y + y' = 1 - e^y$ $y'(xe^y + 1) = 1 - e^y$ $y' = \frac{1 - e^y}{xe^y + 1}$