Calculus AP

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implicit differentiation

Find the slope of the tangent line to the curve xy^3 + 2y - 0.76 =0 at the point (-3, o.7)

I'm lost.
so 3xy^2 * y' + 2y = 0 right?

  • Calculus AP -

    not quite.

    xy^3 + 2y - 0.76 =0
    y^3 + 3xy^2 y' + 2y' = 0
    y' = -y^3/(3xy^2 + 2)

    xy^3 is a product f*g
    (f*g(' = f'g + fg'

    at (-3,0.7) y' = -.7^3/(3(-3)(.7^2) + 2)
    = . . .

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