3x^2+2xy+y^2=2 then the value of dy/dx at x=1 is?

Answer 1

using implicit differentiation,

6x + 2y + 2xy' + 2yy' = 0
when x=1, y=-1, so plug them i:
6 - 2 + 2y' - 2y' = 0
Note that there is no solution, so the tangent to the curve is vertical. y' is not defined.

Looking at dx/dy,
6xx' + 2x + 2yx' + 2y = 0
6x' + 2 - 2x' - 2 = 0
4x' = 0
x' = 0
just as we suspected: dx/dy=0, so the slope is vertical.