## Calculus: Early Transcendentals 8th Edition

$h(x,y)=\dfrac{(1-xy)}{(1+x^2y^2)}+ln[\dfrac{(1-xy)}{(1+x^2y^2)}]$; set of points: {$(x,y) | 1\gt xy$}
We are given: $h(x,y)=\dfrac{(1-xy)}{(1+x^2y^2)}+ln[\dfrac{(1-xy)}{(1+x^2y^2)}]$ The function is continuous everywhere and the $ln$ term is only defined for positive values. Thus, $1-xy\gt 0$ or, $1\gt xy$ Hence, $h(x,y)=\dfrac{(1-xy)}{(1+x^2y^2)}+ln[\dfrac{(1-xy)}{(1+x^2y^2)}]$; set of points: {$(x,y) | 1\gt xy$}