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?
 A: 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).
A: 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:


*

*The unexpanded and expanded denominator are equivalent   

*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.

