Asked by Ines

xy^2 + 2xy = 8, then, at the point (1,2) y' is?
Answered by oobleck
again, use implicit differentiation
y^2 + 2xyy' + 2y + 2xy' = 0
at (1,2) you have
4 + 4y' + 4 + 2y' = 0
y' = -4/3
Answered by Damon
Dear Gang of Bots,
The first reply by oobleck is correct.
Your replies are nonsense.
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