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Using the Quotient Rule with the Chain Rule

  1. Nov 5, 2006 #1
    Hi,

    I'm struggling to understand how to find the derivative of something like this...

    [(2x - 1)^2] / [(x - 2)^3]

    The answer in my book says it is supposed to be [-(2x - 1)(2x + 5)] / [(x - 2)^4]

    How do I use the chain rule with the quotient rule at the same time?

    I even multiplied it all out and tried to find the derivative with just the quotient rule but I can't get the book's final simplifed value.

    Thanks
     
  2. jcsd
  3. Nov 5, 2006 #2
    Remember, that something like 1 / x is equal to x^-1. You can put everything in the numerator, and then use the product rule first.
     
  4. Nov 5, 2006 #3

    radou

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    Homework Helper

    I wrote that a million times before, but I'm too lazy to search. So, let f(x) = (2x -1)^2, g(x) = (x - 2)^3, and the original function h(x) = f(x) / g(x) You have:

    h ' (x) = ( f ' (x)*g(x) - f(x)*g ' (x) ) / ( g(x)^2 ).
     
  5. Nov 5, 2006 #4
    I know that's the quotient rule. My problem is that when I apply it to my problem, I can't seem to get the simplified answer from the book.

    For example, if the denominator is (x - 2)^3 and it's squared, shouldn't that produce (x - 2)^6. Therefore, how could the answer have a denominator of only (x -2 )^4?

    Thanks
     
  6. Nov 5, 2006 #5

    Hurkyl

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    Staff Emeritus
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    Gold Member

    Make an equation out of it.

    [tex]
    \frac{M}{(x-2)^6} = \frac{N}{(x-2)^4}
    [/tex]

    Can this ever be true? When?
     
  7. Nov 6, 2006 #6

    HallsofIvy

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    Science Advisor

    Because you are going to have a factor of at least (x-2)^2 in each term in the numerator and that cancels.
     
  8. Nov 6, 2006 #7
    Thanks a lot. Once I factored out (x-2)^2 from the numerator and denominator, I was able to reduce my answer to what was shown in the book...:smile:
     
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