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
2 answers
-1/6
-1/10