Derivade of: xy = (1 –x –y)2

We'll calculate the partial derivatives of the given
expression:

xy = (1 –x
–y)^2

For the beginning, we'll square raise the right
side:

(1 –x –y)^2 = 1 + x^2 + y^2 - 2x - 2y +
2xy

The expression will
become:

xy = 1 + x^2 + y^2 - 2x - 2y +
2xy

1 + x^2 + y^2 - 2x - 2y + xy =
0

If we'll calculate the derivative with respect to x,
we'll get:

2x +  y^2 - 2 - 2y + y =
0

(d/dx)(1 + x^2 + y^2 - 2x - 2y + xy)' =  2x
+  y^2 - y - 2

Note: the terms in y are
considered constants.

If we'll calculate the derivative
with respect to y, we'll get:

(d/dy)(1 + x^2 + y^2 - 2x -
2y + xy)' = x^2 + 2y - 2x - 2 + x

(d/dy)(1 +
x^2 + y^2 - 2x - 2y + xy)' = x^2 + 2y - x -
2

Note: the terms in x are considered
constants.