Suppose a curve is defined by the equation (x + y) 2 = 1. what is the equation of the line tangent to the curve at (3, - 2) ?

(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

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)