calculus

find the tangent to y=27/(x^2+2) at (1,9)

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  1. y = 27(x^2+2)^-1
    dy/dx = -27(2x)(x^2+2)^-2 = -54x/(x^2+2)^2
    at x = 1 , dy/dx = -54/9 = -6

    equation:
    y = -6x+b , sub in the point (1,9)
    9 = -6+b
    b = 15

    tangent equation: y = -6x + 15

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