Answer
(a) Graph III
(b) Graph I
(c) Graph IV
(d) Graph II
Work Step by Step
(a) $y' = 1+x^2+y^2$
Graph III
When $x =y =0$, the slope is 1.
As $x$ and $y$ both become more positive, the slope increases.
As $x$ and $y$ both become more negative, the slope increases.
(b) $y' = xe^{-x^2-y^2}$
Graph I
When $x=0$, the slope is 0.
When $x \lt 0$, the slope is negative.
When $x \gt 0$, the slope is positive.
(c) $y' = \frac{1}{1+e^{x^2+y^2}}$
Graph IV
When $x =y =0$, the slope is 1.
As $x$ and $y$ both become more positive, the slope decreases.
As $x$ and $y$ both become more negative, the slope decreases.
(d) $y' = sin(xy)cos(xy)$
Graph II
When $y =0$, the slope is 0.
