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

9 years ago

4 years ago

8 months ago

8 months ago

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.

8 months ago

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.

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