calculus

Write an equation of the line tangent to y=(9-x^2)^2/3 at x=1

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  1. dy/dx = (2/3)(9-x^2)^(-1/3) (2x)

    when x = 1, y = 8^(2/3) = 4
    dy/dx = (2/3)(8)^(-1/3) (2)
    = (4/3)((1/2) = 2/3

    equation:
    y = (2/3)x + b
    at( 1,4)
    4 = (2/3) + b
    b = 10/3

    y = (2/3)x + 10/3

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