Find the equation of the tangent line to the curve x^3 + y^3 = 6xy at the point (3,3).

Question

Find the equation of the tangent line to the curve x³ + y³ = 6xy at the point (3,3).

Answer 1

The critical issue here is to differentiate implicitly to get

(3x² + 3y² (dy/dx) = 6[x(dy/dx) + y]

Now collect terms: (3y²-6x) (dy/dx) = y-3x² to get

dy/dx=(6y-3x²)/(3y²-3x²)

Evaluate dy/dx at (3,3) and set that value equal to (y-3)/(x-3) for the equation of the tangent line.