# Using quotient rule to differentiate

Use the quotient rule to differentiate. I want to know if I'm doing this correctly:

$$f(x)=\frac {2x}{x^4+6}$$

First, I find $f$ prime of $x$ and $g$ prime of $x$:

$$f'(x) = 2$$$$g'(x) = 4x^3$$

After using the rule, I end up with:

$$\frac {2x^4+12-8x^4}{(x^4+6)^2} = \frac {-6x^4+12}{(x^4+6)^2}$$

Would this be the final answer if I'm correct? Or do I need to expand the denominator?

• It's fine, except your "$x$" turned into a "$v$". – David Mitra Feb 23 '14 at 21:21
• oh, lol thanks. – o.o Feb 23 '14 at 21:22
• That's absolutely correct, don't expand the denominator! Just, I don't get why you have $v$ instead of $x$? – user88595 Feb 23 '14 at 21:22
• Was a typo. Should I not trust this site: derivative-calculator.net/#expr=%282x%29%2F%28x%5E4%2B6%29? I usually go to it confirm my answers but seems like it is wrong this time. – o.o Feb 23 '14 at 21:24
• That site did it correctly. They used the product rule applied to $2x\cdot{1\over x^4+6}$. Of course, upon simplification, you'll see it's the same as what you obtained. – David Mitra Feb 23 '14 at 21:25

## 2 Answers

As a check, you can use the chain rule (if you've learned it already) and the product rule. (I use it because for a long time I could never remember the order of the quotient rule!)

Let $f(x) = 2x(x^4 + 6)^{-1}$. To find the derivative, apply the product rule: $f'(x) = (2x)'(x^4 + 6)^{-1} + 2x\left[(x^4 + 6)^{-1}\right]'$. Keep going, applying the chain rule:

\begin{aligned} \ 2(x^4+6)^{-1}+(2x)(-1)(x^4+6)^{-2}(4x^3) &= \frac{2}{x^4+6} - \frac{8x^4}{(x^4+6)^2} \\ \ &= \frac{2(x^4+6)-8x^4}{(x^4+6)^2} \\ \ &= \frac{12-6x^4}{(x^4+6)^2} \\ \end{aligned}

It might be a little longer for this problem. However, you can always use it to check your answer on a test (or if you ever forget the quotient rule).

Yes, that's right. Remember that if $h(x) = \frac{f(x)}{g(x)}$, then $h'(x) = \frac{f'(x)g(x) - g'(x)f(x)}{g(x)^2}$. There are two good reasons not to expand the denominator of your final result:

1. The unexpanded and expanded denominator are equivalent
2. The unexpanded denominator can actually be 'easier to work with' for algebraic purposes, so it is better left unexpanded until the need arises to work with its expanded form.