calculus
posted by Taeyeon .
1. If 3x^2+2xy+y2=2 then the value of dy/dx x = 1 is
A. 2 B. 0 C. 2 D. 4 E. not defined

3x^2+2xy+y2=2
6x + 2y + 2xy' + 2yy' = 0
x=1 ==> y=1
6(1) + 2(1) + 2(1)y' + 2(1)y' = 0
6  2 + 2y'  2y' = 0
4 = 0
(e) undefined
The graph is an ellipse. At (1,1) there is a vertical tangent
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