Calculus [finding slope of tangent line]

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

asked by Ali
  1. 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

    posted by Steve

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