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Find derivative of: (3 - x^2)/(4 + x^2); Quotient Rule vs. Product Rule

  1. Jan 10, 2013 #1
    Hi, I got the right answer when I used the Quotient Rule but not when I used the Product Rule...
    I think it might be an algebra mistake...

    Product Rule Method:

    f'(x) = (3 - x^2)*(4 + x^2)^-1

    = (3 - x^2)[(-1(4 + x^2)^-2)*2x] + [(4 + x^2)^-1](-2x)

    = [(3 - x^2)(-2x)]/[(4 + x^2)^2] + [(-2x)/(4 + x^2)]

    To get the same denominator here, I thought I might square the factor on the right.

    = (-6x + 2x^3 + 4x^2)/[(4 + x^2)^2] <----- This is not right.

    THE RIGHT ANSWER IS: -14x/[(4 + x^2)^2]
    I found that by using the Quotient Rule.
    ...But where am I going wrong with the Product Rule...?

    P.S. Oh, and a side note:
    If I wrote out the Product Rule answer like this: (-6x + 2x^3 + 4x^2)/(x^4 + 8x^2 + 16)

    Would the 4x^2 and the 8x^2 be able to cancel somewhat?
    Sometimes I get confused about what can cancel.
    Could it turn into:
    (-6x + 2x^3)/(x^4 + 2 + 16) = (-6x + 2x^3)/(x^4 + 18)?

    I know that's not the right answer, but is this how it would cancel?

    Thank you very much!
  2. jcsd
  3. Jan 10, 2013 #2


    Staff: Mentor

    given a fraction like (a+b / (a + c) you can't cancel the a terms out to get b / c

    but if you have a * b / a * c then you can cancel the a factors out to get b / c

    with that in mind try reviewing your work to see where you went wrong, the product and quotient rules should give the same answers.
  4. Jan 10, 2013 #3


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    Gold Member

    rewritten a bit more correctly what you surely meant

    As far as I understand what you are saying, there is just no rationale for your operation of squaring. You have a sum, represent it X + Y . You are saying if you square a bit of it -> X + Y2 that will be the same thing?

    You have something of form A/B2 + C/B .

    In general how do you combine that into one?
  5. Jan 10, 2013 #4


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    Be careful to write what you mean! This is f, not f'. You have not yet differentiated.
    [itex]f(x)= (3- x^2)(4+ x^2)^{-1}[/itex]

    Okay, this is correct.

    Yes, it is not right. If you have 2/3 and want to make the denominator 9, you multiply both numerator and denominator to get 6/9 which, of course, represents the same number as 2/3. If you square, you get 4/9 which is NOT the same.

    Going back to
    [tex]-\frac{2x(3- x^2)}{(4+x^2)^2}- \frac{2x}{4+ x^2}[/tex]
    if you multipy both numerator and denominator of the second fraction by [itex]4+ x^2[/itex] you get
    [tex]-\frac{6x- 2x^3}{(4+ x^2)^2}- \frac{8x+ 2x^3}{(4+ x^2)^2}[/tex]
    [tex]= \frac{-6x+ 2x^3- 8x- 2x^3}{(4+ x^2)^2}= \frac{-14x}{(4+x^2)^2}[/tex]

    Last edited by a moderator: Jan 10, 2013
  6. Jan 10, 2013 #5
    WOW! HallsofIvy! You gave me the BEST answer I have ever gotten in my experience on Physics Forums!

    You wrote out EXACTLY where I went wrong, and explained WHY it was wrong!

    I get it now! My squaring method was off!

    I worked the problem by multiplying instead by (4 + x^2) and got the right answer!

    Thank you SO much for your help!
    With the other answers, I didn't understand what was wrong with my method!

    If this were YahooAnswers, you'd get voted "Best Answer!" :)
  7. Jan 10, 2013 #6
    Just apply what you've learned: [tex]\frac{(3-x^{2})'(4+x^{2})-(4+x^{2})'(3-x^{2})}{(4+x^{2})^{2}}=\frac{-14x}{(x^{2}+4)^{2}}[/tex]
  8. Jan 11, 2013 #7


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    You heard it here first. :tongue:

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