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Calculus: Limits

  1. Jun 28, 2015 #1
    1. The problem statement, all variables and given/known data
    Find f ' (x) if f(x) = 4x + 4 / x2 + 4

    2. Relevant equations
    I used the mnemonic "lo dhi - hidlo / (lo)^2

    3. The attempt at a solution
    I got -4x^2 +16-8x / (x^2+4)^2
    but it's telling me I'm wrong? Why? I computed it again but I still got the same answer.
  2. jcsd
  3. Jun 28, 2015 #2


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    If you're doing calculus, you need to be able to write your expressions correctly. Please put brackets where they are required. It's impossible to know what expressions you are actually dealing with here.
  4. Jun 28, 2015 #3

    Problem: find f ' (x) if f(x) = (4x + 4) / (x^2 + 4)

    My answer (which was marked wrong by wiley): (-4x^2 + 16 - 8x) / [(x^2 + 4) ^2]

    Is that good enough?
  5. Jun 28, 2015 #4


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    Your answer looks correct to me.
  6. Jun 28, 2015 #5


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    At first I got a different answer, it was because I didn't put brackets around the second term. It's quite easy to make that mistake, so I'll bet Wiley wanted the 8x to be positive (which wouldn't be correct).
  7. Jun 28, 2015 #6


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    Who is this "Wiley" person and how did he or she mark your answer incorrect? If you are using some "mechanical" scoring, those things are notorious for marking wrong anything that is not in exactly the form it wants.
  8. Jun 28, 2015 #7


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    The quotient rule is quite ugly to use in general (which is what you have used to find the answer).

    It is actually much easier to re-write the expression as:

    $$f(x) = \frac{4x + 4}{x^2 + 4} = (4x + 4)(x^2 + 4)^{-1}$$

    This allows you to take advantage of the product and chain rules, and usually you will be able to find the derivatives of quotients much faster:

    $$f(x) = \frac{4x + 4}{x^2 + 4} = (4x + 4)(x^2 + 4)^{-1} = (4)(x^2 + 4)^{-1} - (4x + 4)(x^2 + 4)^{-2}(2x)$$
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