Calculus question

Question

Find an equation of the tangent line to the curve 3xy²-2(x+y)³=4 at the point (2, -1)

Answer 1

Use implicit differentiation.

 

[3y² + 6xyy'] - 6(x + y)²(1 + y') = 0

 

3y² + 6xyy' - 6(x + y)²(1 + y') = 0

 

 

Now isolate y'.

 

3y² + 6xyy' - 6(x + y)² - 6(x + y)²y' = 0

 

6xyy' - 6(x + y)²y' = -3y² + 6(x + y)²

 

 

Factor out y'.

 

y'(6xy - 6(x + y)²) = 6(x + y)² - 3y²

 

 

Divide both sides of the coefficient of y'.  You will then have your derivative.  Then evaluate y' when x=2 and y=-1.  You will then have your slope of the tangent line.

 

Then use the point-slope form of the line to get the tangent line.

 

y = m(x - x₁) + y₁

 

y = y'(x - 2) - 1