# Finding the derivative using quotient rule...

$$\frac{\text{d}}{\text{dt}}\dfrac{2(t+2)^2}{(t-2)^2}$$

I applied the quotient rule:

$$\dfrac{[2(t+2)^2]'(t-2)^2-2(t+2)^2[(t-2)^2]'}{(t-2)^4}$$

$$\dfrac{4(t+2)(t-2)^2-2(t+2)^22(t-2)}{(t-2)^4}$$

$$\dfrac{4(t+2)(t-2)-4(t+2)^2}{(t-2)^3}$$

This was part of a problem where I needed to find the second derivative of a parametric curve but I am stuck on finding this derivative. I typed this problem into wolfram alpha and it gave me $\dfrac{-(16 (x+2))}{(x-2)^3}$, I've been working on this problem for the past hour and can't figure out what I am doing wrong, could someone please explain how to do this?

• Your answer is the same as Wolframs --- either factor out your numerator and rearrange, or else multiply out both to see they're the same. Dec 20, 2013 at 6:40

$$\frac{(t-2)^2 \cdot 4(t+2) - 2(t+2)^2 \cdot 2(t-2)}{(t-2)^4}$$ $$=\frac{(t-2)(t+2) \cdot 4(t-2) - 4(t+2)}{(t-2)^4}$$ $$= \frac{(t+2) \cdot (4t - 8 - 4t - 8)}{(t-2)^3}$$ $$= \frac{-16(t+2)}{(t-2)^3}$$

• Thank you for rewriting the steps, it really helped me!
– Kot
Dec 20, 2013 at 6:47
• No problem, glad to help.
– Zhoe
Dec 20, 2013 at 6:48

Note that $4(t+2)(t-2)-4(t+2)^2=-16(t+2).$ Hence, you are correct.

Write $\displaystyle\frac{t+2}{t-2}$ as $$1 + \frac{4}{t-2}~,$$ and use the chain rule.

Thus

\begin{align*} \frac{\text{d}}{\text{dt}}\dfrac{2(t+2)^2}{(t-2)^2} & = 2\frac{\text{d}}{\text{dt}}\left(1 + \frac{4}{t-2} \right)^2 \\ & = 4\left(1 + \frac{4}{t-2}\right) \left(\frac{-4}{(t-2)^2}\right)\\ & = -\frac{16(t+2)}{(t-2)^3}~. \end{align*}

$$4(t+2)(t-2)-4(t+2)^2 = 4t^2-16 - 4t^2 - 16t -16= -16(t+2).$$