# Derivade of: xy = (1 –x –y)2

### 2 Answers | Add Yours

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.

xy = (1-x-y)^2.

To find the derivative of xy = (1-x-y)^2.

Differentiating both sodes with respect to x we get:

(xy)' = {(1-x-y)^2}'.

xy' +x'y = 2(1-x-y)^(2-1) * (1-x-y)'

xy'+1*y = 2(1-x-y)(0-1-y')

xy' +y = -2(1-x-y) - 2(1-x-y)y'

We solve for y':

xy'+2(1-x-y)y' = -2(1-x-y) -y = -2+2x+2y-y

y' (x+2-2x-2y) = 2x+y-2

y'(2-x-2y) = (2x+y-2)

y' = (2x+y-2)/(2-x-2y) .