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

1 + xy'+y+4y y' = 0

y' = -1-y/(x+4y)

Answer 2

0

Answer 3

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

Answer 4

i think your derivative is off a bit... but it could be me

Answer 5

x; 1
xy; xy' + y
2y^2; 4y y'
=
6; 0

1 +xy' +y +4y y' = 0

xy' +4y y' = -1-y

y'(x+4y) = (-1-y)

y' = (-1-y)/(x+4y) ; at (x=2,y=1)

y'= -1-(1)/..... = 0

Answer 6

your product rule on (xy) is a little off

Answer 7

and i cant seem to add lol

-1-1 = -2
=-1/3
2+4 = 6