Question

Consider the curve x+xy+2y^2=6. the slope of the line tangent to the curve at the point (2,1) is?

Answer 1

the derivative is actually (-y-1)/(4y+x)
then you plug in the points (2,1) ==>
(-1-1)/(4*1+2) which equals -1/3
then you just plug in your info into a point-slope formula and get:
y-1=-1/3(x-2) which can then be reduced

Answer 2

The derivative of your function is:
1 + (1*dy/dx + y) - 4y(dy/dx) = 0
2 + dy/dx + y - 4y(dy/dx) = 0
(2 + y) + dy/dx (1 - 4y) = 0
(dy/dx) = -(2 + y)/(1 - 4y)

and at point (2,1),
dy/dx = -(2 + 1)/(1 - 4*1)
= -3 / -3
= 1

Slope of tangent line = 1