Simplifying a function to get a tangent line

Question

In the middle of solving this implicit function/using implicit differentation to find its tangent line, this is what I have so far:

 

Original function: y^3+x^3-2xy=8 (find tangent line at (2,2))

 

I started the implicit differentation to find y' and got: (3y^2)(y')+3x^2-2xy'+2y=0

 

I know I need to factor out the y' and get it on the LHS but I'm having trouble getting beyond this point correctly...not sure if it's a simple algebraic issue or what.

 

Thank you!

Answer 1

y³ + x³ - 2xy = 8

 

3y²y' + 3x² - 2y - 2xy' = 0

 

(3y² - 2x)y' = 2y - 3x²

 

y' = (2y - 3x²) / (3y² - 2x)

 

We verify that (2,2) is on the curve.

 

2³ + 2³ - 2·2·2 = 8 + 8 - 8 = 8 Yes.

 

The slope at (2,2) is

 

y'(2,2) = (2·2 - 3·2²) / (3·2² - 2·2) = -1

 

Tangent line:

 

Point-slope form: y = -1(x - 2) + 2

 

Slope-intercept form: y = -x + 4