calculus

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

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