slope question?

Question

consider the curve of x+xy+2y^2=6. The slope of the tangent line to the curve at the point (2,1) is
a)2/3
b)1/3
c)-1/3
d)-1/5
e)-3/4

Answer 1

x+xy+2y^2=6
Solve using the product rule of derivatives with respect to y:
Product rule states: f'[x]g[x] + f[x]g'[x]

1 + (y+xy')+4y*y'=0
1 + y+xy'+4y*y'=0
-1-y= xy'+4y*y'
-1-y= y'(x+4y)
-1-y/x+4y= y'

At point (2,1),
-1-y/x+4y= y'
-1-(1)/(2)+4(1)= y'
-2/6=y'
-1/3=y'=slope

[Answer: C]

Answer 2

c) - 1/3 is the answer worked out as under.

x + xy + 2y^2 = 6
=> 1 + y + xy'+ 4yy'= 0
Plug (2, 1) to get
1 + 1 + 2y'+ 4(1)y'= 0
=> 6y'= - 2
=> slope, y' = - 1/3

Answer 3

1.Take the derivative of the line.
1+(y+y'x)+4yy'=0
2. solve for y'
y'(x+4y)=(-1-y)
y'=(-1-y)/(x+4y)
3.plug in x and y
y'=(-1-1)/(2+4*1)=-2/6=-1/3
The slope is negative one over three.

Answer 4

u in calculus?

ur slope formula is this....

1+ (y + y'x)+(4y . y') =0


work it from there......and then find y' and plug (2,1) in for x.

hey i'm gettin -1/3.......

Answer 5

b