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Homework Help: Problem - find second derivative

  1. Dec 15, 2012 #1
    The problem and my attempt at a solution is shown in the attached image.

    The problem is that I end up with one extra x in the denominator.
    So the question is: Is my expression for y'' correct and I just made a mistake somewhere (I have checked it several times), or am I missing something in the expression for y' and/or y'' ?


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  2. jcsd
  3. Dec 16, 2012 #2


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    If the answer in the book is something like 6/x^3 + 18/x^4 + 12/x^5, you got the same answer.
  4. Dec 16, 2012 #3


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    How do you go from [itex]\displaystyle\ \frac{6}{x^3}\left( 1+\frac{3}{x}+\frac{2}{x^2}\right)\ [/itex] to [itex]\displaystyle\ \frac{6}{x^4}\left( 1+\frac{1}{x}\right)\left( 1+\frac{2}{x}\right)\ ? [/itex]

    [itex]\displaystyle\ 1+\frac{3}{x}+\frac{2}{x^2}=\left( 1+\frac{1}{x}\right)\left( 1+\frac{2}{x}\right)\ . [/itex]
  5. Dec 16, 2012 #4


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    Does this make sense?
    [tex] \[
    y' = 3(1 + {\textstyle{1 \over x}})^2 x^{ - 2} ( - 1) \\
    = \frac{{ - 3(1 + {\textstyle{1 \over x}})^2 }}{{x^2 }} \\

    Next you would use quotient rule to find the next derivative (second derivative).
  6. Dec 16, 2012 #5
  7. Dec 16, 2012 #6


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    Personally, I'd just expand the binomial then differentiate twice. Keep the exponents as negative powers, it makes everything quicker.
  8. Dec 16, 2012 #7
    >>Personally, I'd just expand the binomial then differentiate twice
    That certainly makes it much easier. I guess I didnt think of that since I was doing problems related to the chain rule. Thank you for your suggestion.
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