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If a given first derivative is: $\ {dy \over dx} = {-48x \over (x^2+12)^2} $

What are the steps using the quotient rule to derive the second derivative: $\ {d^2y \over dx^2} = {-144(4-x^2) \over (x^2+12)^3} $

My Steps:

\begin{align*} {d^2y \over dx^2} &= {-48(x^2 +12)^2 - 2(x^2+12)(2x)(-48x)\over (x^2+12)^4} \\ &=-48{(x^2 +12)^2 - 4x^2(x^2+12)\over (x^2+12)^4} \\ &= -48{(x^2 +12)( - 4x^2 +(x^2+12))\over (x^2+12)^4} \\ &= -48{(x^2 +12)( - 4x^2)\over (x^2+12)^3} \\ &= {???} \end{align*}

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Where are you getting stuck? – dfeuer Oct 27 '13 at 17:27
Writing down the steps now. Will edit in a moment. – user40650 Oct 27 '13 at 17:27
^that's the spirit! (+1) – The Chaz 2.0 Oct 27 '13 at 17:32
Edited to show where I am getting lost at. – user40650 Oct 27 '13 at 17:42
Just saw this. Are you still stuck? – The Chaz 2.0 Oct 27 '13 at 19:01

3 Answers 3

up vote 4 down vote accepted

We can factor out the constant term to make life easier (and just multiply by that $-48$ at the end of our calculation) as:

$$-48 \dfrac{x}{(x^2+12)^2}$$

This makes it easier to use the quotient and chain rule (I will assume you know these).

The derivative of $\ {dy \over dx} = {-48x \over (x^2+12)^2} $, using the quotient and chain rule is:

$\dfrac{d^2y}{dx^2} = -48 \dfrac{(1)(x^2+12)^2 - 2(x^2+12)(2x)x}{(x^2+12)^4} = -48 \dfrac{(x^2+12) - 2(2x)x}{(x^2+12)^3} = \dfrac{-144(4-x^2)}{(x^2+12)^3}$

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I am getting up to the second portion of your solution. Where does the -144 come from? – user40650 Oct 27 '13 at 17:37
You are also using another rule, by factoring out the $-48$. Any chance you'd care to show where each rule comes into play? Obviously everyone who already knows how to do this calculation will have no trouble following along, but the OP seems lost. – The Chaz 2.0 Oct 27 '13 at 17:37
Factor an $(x^2+12)^1$ from the numerator and cancel it with the denominator. Clear? – Amzoti Oct 27 '13 at 17:38
@user40650: Are you following this and got it? – Amzoti Oct 27 '13 at 17:46
Your problem is the $-48x$ term, that should be multiplied, not subtracted. For the second equality just do the factor of $(x^2+12)$ from the numerator and cancel with denominator. Got it? – Amzoti Oct 27 '13 at 17:53

First, make sure you know what the quotient rule is - which you can derive from the product rule say by taking the derivative of $g(x)/f(x)$.

Then, simplify. Looking at the product rule, expect to have to cancel a factor of $x^2+12$ from the top and bottom.

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Re: the now-deleted answer to this question. I agree that we should not spoon-feed answer. However, "should" represents a pedagogical philosophy that is far from universal. – The Chaz 2.0 Oct 27 '13 at 17:35

Recall the Quotient Rule:

$$(\frac{f}{g})' = \frac{f'\cdot g - g' \cdot f}{g^2}$$

So you have the first derivative, $\frac{dy}{dx}= \frac{-48x}{(x^2+12)^2}$. Just apply the quotient rule:

Let $f(x) = -48x$. So $f'(x) = -48$.
Let $g(x) = (x^2 + 12)^2$. So, by the chain rule, $g'(x) = 2(x^2 + 12) \cdot 2x = 4x(x^2+12)$.

$$\frac{d^2y}{dx^2} = \frac{(-48 \cdot (x^2+12)^2) - (4x(x^2+12) \cdot -48x)}{((x^2+12)^2)^2}$$ Notice that both elements of the numerator contain $-48$. We can factor that out. $$\frac{d^2y}{dx^2} = \frac{-48 \cdot ((x^2+12) - (4x^2(x^2+12)))}{(x^2+12)^4}$$ We also can factor out the $(x^2+12)^2$ term: $$\frac{d^2y}{dx^2} = \frac{-48 \cdot (x^2+12) \cdot ((x^2+12) - 4x^2)}{(x^2+12)^4}$$ And now we can cancel: $$\frac{d^2y}{dx^2} = \frac{-48 \cdot (12 - 3x^2)}{(x^2+12)^3}$$ Observe that we can factor the numerator some more: $$\frac{d^2y}{dx^2} = \frac{-48 \cdot 3 \cdot (4 - x^2)}{(x^2+12)^3}$$ And there we are: $$\frac{d^2y}{dx^2} = \frac{-144 \cdot (4 - x^2)}{(x^2+12)^3}$$

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Your denominator is funky funky. – dfeuer Oct 27 '13 at 17:31
My impression is this kind of question should be answered with hints - not a direct answer. – levitopher Oct 27 '13 at 17:31
@dfeuer embarrassing mistake; fixed. – Newb Oct 27 '13 at 18:07
@levitopher Possibly, but I try not to read too much into the question asked by the person. Above all, I try to provide an informative and clear answer. – Newb Oct 27 '13 at 18:09

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