calculus PLZ HELP!!
posted by Liz .
find the equation for the line tangent to the curve at the given point.
(x^2)(y^2)2xy+x=1 at (2, 1/2)

(x^2)(2y)dy/dx + 2x(y^2)  2xdy/dx  2y +1 = 0
dy/dx(2x^2 y  2x) = 2x y^2 + 2y  1
dy/dx = (2xy^2 + 2y  1)/(2x^2y  2x)
at (2, 1/2)
dy/dx = (2(2)(1/4) + 2(1/2)  1)/(2(4)(1/2)  4)
= (1 + 1 1)/0
ahh ,we are dividing by zero, so the tangent must be a vertical line
since it passes through (2, 1/2) its equation must be
x = 2
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