Understanding Parametric Second Derivative through Polynomial Division

[Andrusko]

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:

$$\frac{\frac{d}{dt}(\frac{3t^{2}-3}{3t^{2}-6t})}{3t^2-6t}$$

and is meant to end up as:

$$\frac{-2(t^{2}-t+1)}{3t^{3}(t-2)^{3}}$$

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.

$$\frac{u}{v} = \frac{u'v-uv'}{v^{2}}$$

My rearrangement looks like this:

$$\frac{-3t^{2}-t+1}{(3t^{2}-6t)^{3}}$$

 

[Gib Z]

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

$$\frac{u}{v} = \frac{u'v-v'u}{v^2}$$ is incorrect, although

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

Firstly, I would have done polynomial division to get $$\frac{3t^2-3}{3t^2 -6} = 1 + \frac{2t-1}{t^2-2t}$$ 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.

 

[Andrusko]

But it's $$3t^{2}-6t$$ on the bottom, not $$3t^{2}-6$$

Can you still divide them?

[edit] Oops turn out I just suck at polynomial division. I got it now... [edit]