# calculus PLZ HELP!!

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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)

• calculus PLZ HELP!! -

(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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