Derivatives

Question

Find the equa of the tangent line at point (2,3) on the function x^3+y^3=6xy-1

 

 

 

my equation I got doesn't graph correctly please help!

Answer 1

using implicit differentiation 3x² + 3y²•y' = 6xy' + 6y which leads to y' = (2y-x²) / (y²-2x).

 

y'(2,3) = 2/5 gives us the slope of the tangent.

 

Using this slope and the Point Slope form, you can write the equation of the tangent line now.