If y is a differentiable function of x, then the slope of the tangent to the curve xy - 2y + 4y^2 = 6 at the point where y=1 is ......
How do I do this?
Thanks.
xy - 2y + 4y^2 = 6 --->
d[xy - 2y + 4y^2] = 0 --->
xdy + ydx - 2dy + 8ydy = 0 --->
(x + 8y - 2)dy/dx = -y ---->
dy/dx = y/[2-x-8y]
You know what x is for y = 1 from the equation:
xy - 2y + 4y^2 = 6
-1/10
Well, solving the equation xy - 2y + 4y^2 = 6 for y = 1 will give you the x-coordinate of the point where y = 1 on the curve. Once you have that x-coordinate, you can plug it into the expression dy/dx = y/[2-x-8y] to find the slope of the tangent at that point.
To find the slope of the tangent to the curve at the point where y = 1, you need to substitute y = 1 into the equation to solve for x. Let's do that:
xy - 2y + 4y^2 = 6
Substituting y = 1:
x(1) - 2(1) + 4(1)^2 = 6
x - 2 + 4 = 6
Simplifying further:
x + 2 = 6
x = 4
Now that you know x = 4 when y = 1, you can substitute these values back into the equation:
dy/dx = y/[2-x-8y]
dy/dx = 1/[2-4-8(1)]
dy/dx = 1/[-10]
dy/dx = -1/10
Therefore, the slope of the tangent to the curve at the point where y = 1 is -1/10.
To find the slope of the tangent to the curve at the point where y = 1, we can substitute y = 1 into the equation:
x(1) - 2(1) + 4(1^2) = 6
Simplifying this equation gives us:
x - 2 + 4 = 6
x + 2 = 6
x = 6 - 2
x = 4
So, when y = 1, x = 4.
Now, we can substitute these values into the derivative we found earlier:
dy/dx = y/[2-x-8y]
dy/dx = 1/[2-4-8(1)]
dy/dx = 1/[-10]
dy/dx = -1/10
Therefore, the slope of the tangent to the curve at the point where y = 1 is -1/10.