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?
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
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.
To find the third derivative of the function f(x) = x(3x-7)^3, we can use the chain rule and the power rule of differentiation.
Let's break down the process step by step:
Step 1: Expand the function using the binomial theorem. The function can be rewritten as:
f(x) = x(27x^3 - 189x^2 + 441x - 343)
Step 2: Apply the product rule to find the first derivative:
f'(x) = 1 * (27x^3 - 189x^2 + 441x - 343) + x * (81x^2 - 378x + 441)
= 27x^3 - 189x^2 + 441x - 343 + 81x^3 - 378x^2 + 441x
= 108x^3 - 567x^2 + 882x - 343
Step 3: Apply the product rule again to find the second derivative:
f''(x) = 3 * 108x^2 - 2 * 567x + 882
= 324x^2 - 1134x + 882
Step 4: Finally, apply the product rule for the third time to find the third derivative:
f'''(x) = 2 * 324x - 1134
= 648x - 1134
From the above steps, we can see that the correct answer for the third derivative is 648x - 1134. Therefore, if you are consistently getting 648x - 1134 while attempting the problem, then it seems you are doing it correctly.