Use implicit differentiation to find an equation of the tangent line to the curve at the given point.

Question

x² + 2xy − y² + x = 12,  (3, 6)

  (hyperbola)

Answer 1

f(x) = x^2 + 2xy - y^2 + x = 12

f'(x) = 2x + 2xy' + 2y - 2yy' + 1 = 0

y'(2x - 2y) + 2(x + y) + 1 = 0

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

at (3,6)

y' = (-2(9) + 1)/(-6) = -17/(-6) = 17/6

y - y1 = m(x=x1)

y - 6 = (17/6)(x-3)

y = (7/16)x + 77/16

Answer 2

Differentiating both sides of equation with respect to x (d/dx) and keeping in mind the proper application of the chain rule, we get:

2x + (2y + 2x (dy/dx)) - 2y(dy/dx) + 1 = 0

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

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

at (3,6), m = (-1-6-12)/(6-12) = -19/-6 = 19/6

y = mx + b

6 = (19/6)(3) + b

Solve for b.