Question

For the curve defined implicitly by x^3 + y^2−2xy=2, find the slope of the tangent line at (-1,1). Write your answer either as an integer or a reduced fraction.

Answer 1

-5/2

Explanation:

Answer 2

Explanation:

The method you use is sort of a chain rule when you hit a y. Here's what I get

d(x^3)/dx = 3x^2

d(y^2)/dx = 2y * (dy/dx) This is where the chain rule comes in.

-d(2xy)/dy = -2y - 2x dy/dx

d(2)/dx = 0 Two is a constant. Its derivative = 0.

Now add these together.

3x^2 + 2y*(dy/dx) - 2y - 2x*(dy/dx) = 0

Now group the terms not containing dy/dx

3x^2 - 2y

And then group those containing dy/dx

2y (dy/dx) - 2x (dy/dx)

Pull out the common factor 2[dy/dx]

2 (dy/dx) [y - x]

Put the non dy/dx terms on the right

2(dy/dx) ( y + x) = - (3x^2 + 2y)

Divide both sides by 2(y - x)

(dy/dx) =
`(-(3x^2+ 2y))/(2(-x + y))`

You have the tangent expression. dy/dx is one expression for the tangent. The other is the fraction on the right.

You are almost home free.

Let x = - 1

Let y = 1

dy/dx = -(3(-1)^2 + 2(1) ) / (2(1 + 1)

dy/dx = -(3 + 2) /(2*2)

dy/dx = - (5)/4

dy/dx = - 1.25

Check my numbers pretty carefully. The method is correct. There are so many lines that the answer may not be.