Find an equation of the tangent line at the given point.

Question

xy+x^2y^2=6

(2,1)

Answer 1

Patrick B. has used the method of implicit differentiation to find the value of slope to be -1/2.

To find the equation of the tangent line, continue with

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

y = -(1/2) x + 2

Answer 2

xy' + y + 2x^2 yy' + 2xy^2 = 0

xy' + 2x^2 yy' = -2xy^2 - y

y' ( x+2x^2y) = -2xy^2 - y

y' = (-2xy^2 - y )/( x + 2x^2y)

= -y( 2xy + 1)/ { x( 1 + 2xy) }

= -y/x , as 2xy+1 cancels provided that xy is NOT -1/2

(2,1) ---> y' = -1/2