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altavistard
Don't you mean (x + y)^2 = 1?

Take the derivative with respect to x of both sides of this equation:

2(x + y(dy/dx) = 0.  Then y(dy/dx) = -x, and (dy/dx) = -x/y.

Thus, the slope of the tangent line to this curve at (3, -2) is m = -3 / (-2), or 3/2.

The eqn of the T. L.   is then   y-(-2) = (3/2)(x-3) (answer)
(x + y)^2 = 1

Take the derivative of both sides wrt x.
d/dx (x + y)^2 = 0
d/dx (x^2 + 2xy + y^2) = 0
2x + 2y + 2x*dy/dx + 2y*dy/dx = 0

dy/dx (2x + 2y) = -2x - 2y
dy/dx = (-x - y)/(x + y)
dy/dx = -(x + y)/(x + y) = -1

So anywhere on the curve (x + y)^2 will have a tangent of -1
Eqn of tangent: y +2 = -(x - 3)
y = 3 - x - 2
y = 1 - x

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