# 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

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}
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: