calculus help

find y' and y'' for the equation below.

Show work please!!!

y= (x^3)/((x+1)^2)

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  3. 👁 178

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  2. [(x+1)^2 (3 x^2)-x^3 (2)(x+1)] /(x+1)^4

    [(x+1)(3x^2) - 2 x^3 ]/ (x+1)^3

    [ 3 x^3 + 3 x^2 -2 x^3 ] / (x+1)^3

    [ x^3 + 3 x^2] /(x+1)^3

    x^2 (x+3) / (x+1)^3

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  3. then do it again :)

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    6x / (x+1)^4

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