Find the slope of the tangent line to the curve 2x^2+1xy–4y^3=–18 at the point (6,3)?

Answer 1

you have to find the derivative implicitly:
4x + y + x dy/dx - 12y² dy/dx = 0

4x+y = 12y² dy/dx - x dy/dx

4x + y = dy/dx (12y² - x)
dy/dx = (4x+y)/(12y²-x)

So the derivative, slope of tangent line, at (6,3) is

(4*6 + 3) / (108 - 6)

=27/102
= 9/34

Answer 2

Let y' = dy / dx

Implicitly differentiate with respect to x:

2x^2+ xy – 4y^3 = -18

4x + xy' + y - 12y^2 y'= 0

Solve for y':

y'(x - 12y^2) = -4x - y

y' = (-4x - y) / (x - 12y^2)

Evaluate y' at (x, y) = (6, 3):

y' = (-4(6) - (3)) / ((6) -12(3)^2)

y' = -27 / (-102)

y' = 9 / 34

Answer: The slope of the tangent line to the curve 2x^2+1xy–4y^3=–18 at the point (6,3) is 9 / 34.

Answer 3

Hrmm.
Let's see if we can get the implicit derivative here...
2x^2+xy - 4y^3 =18
4x + y +xy' - 12y^2 y'= 0
x=6, y=3
24 + 3 + 6y' -108y' = 0
27 - 102y' = 0
-102y' = -27
y' = 27/102 = 9/34
That help?
I think I did that right.

Answer 4

Implicit differentiation gives
4x + (y + x dy/dx) - 12y^2 dy/dx = 0
<=> dy/dx (x - 12y^2) + (4x + y) = 0
<=> dy/dx = -(4xy) / (x - 12y^2)
At the point (6, 3) we have
dy/dx = -(4.6.3) / (6 - 12.9) = 12/17.