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Find the slope of the tangent line to the curve 0x^2−2xy+2y^3=10 at point (-4,1)

Help please? Thanks in advance

1 answer

Don't know why you bother with the zero:
−2xy+2y^3=10
(yes, (-4,1) is on that graph.

just use the product rule, remembering that x' = 1:

-2yx' - 2xy' + 6yy' = 0
-2y + (6y-2x)y' = 0

y' = 2y/(6y-2x)
at (-4,1),
y' = 2(1)/(6(1)-2(-4)) = 2/(6+8) = 1/7
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