Answer
${\triangledown}f = (x-y)i + (y-x)j$
Gradient vector field sketch:
Work Step by Step
$f(x,y)=\frac{1}{2}(x-y)^2$
Using the chain rule of differentiation we have:
$f_{x}(x,y) = (1)(x-y)$
$f_{y}(x,y) = (-1)(x-y)$
Therefore, the gradient vector field of $f$ is:
$(x-y)i + (y-x)j$
Sketch of the gradient vector field:
