x^2 + xy - y^3 = xy^2

Question

Find the equation of the tangent line to the curve at the point (1,1)

Answer 1

in order to find the equation of the tangent line you need to slope which is equal to the derivative. use implicit differentiation

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

solve for dy/dx

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

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

then dy/dx = [-2x - y + y^2]/ [x - 3y^2 - 2xy]

now to find the value of the slope use your point (1,1) for x and y

dy/dx = [-2(1) - 1 + 1^2]/ [1 - 3(1^2) - 2(1)(1)]

dy/dx = -2/-4  so dy/dx = 1/2

now use point-slope with m=1/2 and the point (1,1)

y - 1 = 1/2(x -1)

y - 1 = 1/2x - 1/2

y = 1/2x + 1/2  this is the equation of your tangent line.