Question

what is the equation of a line tangent to the curve x^3+y^3=6xy at point (3,3) using implicit differentiation

Answer 1

x^3+y^3=6xy



3x^2+3y^2*y'=6y+6xy'



3y^2*y'-6xy'=6y-3x^2



y'(3y^2-6x)=6y-3x^2



y'=(6y-3x^2)/(3y^2-6x)



y'=(3(2y-x^2))/(3(y^2-2x))



y'=(2y-x^2))/(y^2-2x)



y'=(2(3)-(3)^2)/((3)^2-2(3))



y'=-3/3



y' = -1



So the slope of the tangent line is -1



y = mx+b


3 = (-1)(3)+b


3 = -3 + b


6 = b


b = 6


So the equation of the tangent line is y = -x+6

Answer 2

alexray19 you forgot to derive 6xy to get 6y+6xy'

Answer 3

Oh you're right, thanks :)

Answer 4

thanks, having to refresh on my calc

Answer 5

A solution using Mathematica, including a solution plot and comments is attached.