Question

using implicit differentiation find the equation of the tangent line to the graph of the following function at the indicated point x^2 y^3 -y^2+xy-1=0 at (1,1)

Answer 1

x^2(3y^2dy) + 2x dx(y^3) - 2ydy +xdy + dx y = 0

dy (3x^2y^2-2y +x) +dx(2xy^3+y)=0

dy (3-2+1) + dx (2+1) = 0

2 dy = -3dx

m = slope = dy/dx = -3/2

so y = -3/2 x + b
1 = -3/2 + b

b = 5/2

so y = -3/2 x + 5/2
or
2 y = -3 x + 5