1
$\begingroup$

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

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – coffeemath Dec 20 '13 at 6:40
1
$\begingroup$

Your answer is correct, Wolfram is just simplified.

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

$\endgroup$
  • $\begingroup$ Thank you for rewriting the steps, it really helped me! $\endgroup$ – Kot Dec 20 '13 at 6:47
  • $\begingroup$ No problem, glad to help. $\endgroup$ – Zhoe Dec 20 '13 at 6:48
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

Thus

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

$\endgroup$
0
$\begingroup$

Your numerator equals WA's numerator:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.