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I know this is a very basic question but I need some help.

I have to find the second derivative of:

$$\frac{1}{3x^2 + 4}$$

I start by using the Quotient Rule and get the first derivative to be:

$$\frac{-6x}{(3x^2 + 4)^2}$$

This I believe to be correct. Following that I proceed to find the second derivative in the same manner but I get this as my answer: $$\frac{(54x^4 + 144x^2 +96) - (-36x^3 + 48x)}{(9x^4 +24x^2 +16)^2}$$

This I believe to be correct just not simplified. However the answer I need to get is: $$- \frac{6(4 - 9x^2)}{(3x^2 + 4)^3}$$ I do not know what the best way to approach this would be, should I multiply out the denominator and try to cancel? Could someone point me in the right direction, I want to solve it myself but I need some guidance.

Thanks

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    $\begingroup$ General advice that is quite useful here: don't open parentheses (distribute) unless you have a reason to. $\endgroup$
    – Sammy Black
    May 15, 2013 at 17:10
  • $\begingroup$ @Jan, look more closely at the denominators. The OP has expanded certain polynomials; whereas, the second answer has them still factored. $\endgroup$
    – Sammy Black
    May 15, 2013 at 17:14

3 Answers 3

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The first derivative is correct. Now we want to differentiate $\frac{-6x}{(3x^2+4)^2}$. The main thing to remember is do not "simplify" unless there is good reason to do so.

The derivative of $\frac{-6x}{(3x^2+4)^2}$ is $$\frac{(3x^2+4)^2 (-6)-(-6x)(6x)(2)(3x^2+4)}{(3x^2+4)^4}.$$ Cancel a $3x^2+4$, and simplify the top.

Remark: I probably would want to take out that ugly $-6$ from the top, which is an invitation to minus sign errors and other errors, and differentiate $\frac{x}{(3x^2+4)^2}$.

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    $\begingroup$ Perfect, you explained it extremely well. Your advice is also greatly appreciated. $\endgroup$
    – user
    May 15, 2013 at 17:28
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When you perform the quotient rule, it's often easier to not multiply everything out until the end, because there are a lot of cases where you can factor things out and cancel, and if you multiply out first, it will be much harder to see that. Your first derivative is indeed correct, but here's what I'd recommend to get the suggested answer for the second derivative: \begin{align*} \frac{d}{dx}\left(\frac{-6x}{\left(3x^2 + 4\right)^2}\right) &= \frac{-6\left(3x^2 + 4\right)^2 - (-6x)(2)\left(3x^2 + 4\right)(6x)}{\left(3x^2 + 4\right)^4}\\ &= \frac{6\left(3x^2 + 4\right)\left(-\left(3x^2 + 4\right) + 12x^2\right)}{\left(3x^2 + 4\right)^4}\\ &= \frac{6\left(9x^2 - 4\right)}{\left(3x^2 + 4\right)^3}, \end{align*} which is what you wanted.

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  • $\begingroup$ Thank you for your help also it is appreciated. $\endgroup$
    – user
    May 15, 2013 at 17:28
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$$\left(-\frac{6x}{(3x^2+4)^2}\right)'=-\frac{6(3x^2+4)^2-72x^2(3x^2+4)}{(3x^2+4)^4}=$$

$$=-\frac{18x^2+24-72x^2}{(3x^2+4)^3}=-\frac{6(-9x^2+4)}{(3x^2+4)^3}$$

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