# 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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