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Differentiate the function (derivatives, chain rule for powers)

  1. May 7, 2013 #1
    1. The problem statement, all variables and given/known data
    Differentiate f(x) = (x^2 - 3x)^2


    2. Relevant equations
    f'(x) = nf'(x)f(x)^(n-1)


    3. The attempt at a solution
    f’(x) = 2(x2-3x)’(x2-3x)2-1
    = 2(2x-3)(x2-3x)1
    = 2(2x3 – 6x2 – 3x2 + 9x)
    = 2(2x3 – 9x2 + 9x)
    = 4x3 – 18x2+ 18x

    Is this correct?
     
  2. jcsd
  3. May 7, 2013 #2

    Fredrik

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    Gold Member

    You need to put more effort into how you type your questions. We shouldn't have to spend more time thinking about how to interpret your notation than you did typing it. Here's a link to the LaTeX guide: https://www.physicsforums.com/showpost.php?p=3977517&postcount=3

    If you need to post a question before you have learned the basics of LaTeX (which should take less than 10 minutes) you should at least use vBulletin's sup tags (also mentioned in the guide).

    What you wrote under relevant equations is nonsense. The notation ##(x^2-3x)'## should never be used. The prime should be on a function, not a number. That's why we write f'(x) instead of (f(x))'. You can write ##\frac{d}{dx}(x^2-3x)## instead.

    Apart from that, your calculation looks OK.
     
  4. May 7, 2013 #3
    That could have been worded less offensively, but thank you.
     
  5. May 7, 2013 #4
    actually, i think he worded it pretty well. the problem statement is worded correctly, you should take the time to use the right notation to make things clear, for instance use ^ when you're raising something to a power so people who are trying to help you understand it..otherwise its hard to understand, plus it looks sloppy. other than that...yeah, i got the same answer as you. pretty straightforward problem. you could also factor out 2x if you wanted, wouldn't make much difference. so (2x)(2x-3)(x-3) could also be your answer :approve:
     
    Last edited: May 7, 2013
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