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Parametric second derivative

  1. Oct 2, 2008 #1
    I'm having trouble seeing how an example comes out because the "worked example" skips about 5 steps and I can't get from point a to b.

    It starts as:

    [tex] \frac{\frac{d}{dt}(\frac{3t^{2}-3}{3t^{2}-6t})}{3t^2-6t} [/tex]

    and is meant to end up as:

    [tex] \frac{-2(t^{2}-t+1)}{3t^{3}(t-2)^{3}} [/tex]

    I end up with a mess looking nothing like that and I suspect it's because I've missed some cunning common factor that is easy to spot provided you've already done the problem.

    I used the quotient rule for the top part.

    [tex] \frac{u}{v} = \frac{u'v-uv'}{v^{2}} [/tex]

    My rearrangement looks like this:

    [tex] \frac{-3t^{2}-t+1}{(3t^{2}-6t)^{3}} [/tex]
  2. jcsd
  3. Oct 2, 2008 #2

    Gib Z

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    Homework Helper

    Well that was a good start, although you shouldn't be tacky with your notation!

    [tex] \frac{u}{v} = \frac{u'v-v'u}{v^2} [/tex] is incorrect, although

    [tex] \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v-v'u}{v^2} [/tex] is correct.

    Firstly, I would have done polynomial division to get [tex] \frac{3t^2-3}{3t^2 -6} = 1 + \frac{2t-1}{t^2-2t} [/tex] which simplifies the algebra you have to grind through considerably.

    Do it one step at a time. Let u = 2t-1, so u' = 2. v = t^2 - 2t, so v' = 2t-2. Substitute those directly into the rule again and simplify. You should get what the book says, the result is correct.
  4. Oct 2, 2008 #3
    But it's [tex] 3t^{2}-6t[/tex] on the bottom, not [tex]3t^{2}-6[/tex]

    Can you still divide them?

    [edit] Oops turn out I just suck at polynomial division. I got it now... [edit]
    Last edited: Oct 2, 2008
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