Calculus

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Find the third Derivative of
f(x)=x(3x-7)^3
The answer is 3x-1134 but I attempted the problem multiple times and I keep ending up with 648x-1134. What am I doing wrong?

  • Calculus -

    rather than using the product rule and repeated simpification, I thought I would expand the whole thing
    f(x) = x(3x-7)^3
    = x[ (3x)^3 + 3(3x)^2 (-7) + 3(3x)(-7)^2 + (-7)^3 ]
    = x[ 27x^3 - 189x^2 + 441x - 343]
    = 27x^4 - 189x^3 + 441x^2 - 343x

    f'(x) = 108x^3 - 576x^2 + 882x
    f''(x) = 324x^2 - 1152x + 882
    f'''(x) = 648x - 1152


    btw, Wolfram agrees with my answer
    I kept feeding the previous derivative back in..
    here is the last step

    http://www.wolframalpha.com/input/?i=18%2818x%5E2+-+63x+%2B+49%29

  • Calculus -

    f = x(3x-7)^3

    f' = (3x-7)^3 + 9x(3x-7)^2

    f'' = 9(3x-7)^2 + 9(3x-7)^2 + 54x(3x-7)
    = 18(3x-7)^2 + 54x(3x-7)

    f''' = 108(3x-7) + 54(3x-7) + 162x
    = 162(3x-7) + 162x
    = 162(4x-7)
    = 648x - 1134

    We have a winner! Book is wrong.

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