Find dy/dx: (x + y)^2 = 4

Hello!

Let's differentiate the given equation:

`((x+y(x))^2)' = 4',`

`2(x+y(x))(x+y(x))' = 0,`

`2(x+y)(1+y') = 0.`

We know that `(x+y)^2 = 4,` therefore `x+y = +-2` and it is never zero. Thus `1+y'(x) = 0` and `y'(x) = -1.` This is the answer (dy/dx and y'(x) are the different notations of the same notion).