Tangent line with implicit differentiation

Question

Solve the problem by typing the missing quantities on the spaces provided.

x^3+y^3=6xy−1 at point (2,3)

Write an equation of the tangent line to the graph of the function given implicitly by the equation  at point (2,3)

Answer: (Fraction must be written as 1/2. Do not use spaces in your answer.)

y - Answer=Answer(x-Answer)

Answer 1

Implicit differentiation to find the slope:

Take the derivative of both sides of the equation:

3x²⁺3y²(dy/dx)=6x(dy/dx)+6y

Get all the dy/dx terms to one side of the equation and everything else to the other side

3x²-6y=6x(dy/dx)-3y²(dy/dx)

Factor out (dy/dx)

3x²-6y=(dy/dx)(6x-3y²)

Divide both sides of the equation by (6x-3y²)

dy/dx = (3x²-6y)/(6x-3y²)

Plug in the given x and y values of (2,3)

dy/dx = (3(2)²-6(3))/(6(2)-3(3)²)

dy/dx = -6/-15 = 2/5

Plug everything in

(y-3)=2/5(x-2)