Finding the derivative of a messy function I need to find the derivative of this function. I think I managed to do this, but the simplification is messy. Is there a good way to keep things organized while I go through all of the needed differentiation rules? It was hard to stay focused in even writing this in latex.
$y = \frac{\sqrt[3]{2x}\sqrt{x+5}}{(4x-1)^9}$ 

numerator(u)/denominator(v)
$u = \sqrt[3]{2x}\sqrt{x+5}$
$v = (4x-1)^9$ 
derivative of numerator(u')/denominator(v')
$u' =\frac{(2x)^{\frac{1}{3}}(x+5)^{-\frac{1}{2}}}{2} + \frac{(x+5)^{\frac{1}{2}}(2x)^{-\frac{2}{3}}}{3}$  product rule, chain rule
$v' = 36(4x-1)^8$  chain rule  

quotient rule $\frac{vu' - uv'}{v^2}$ 
$= \frac{\bigg(\frac{(2x)^{\frac{1}{3}}(x+5)^{-\frac{1}{2}}}{2} + \frac{(x+5)^{\frac{1}{2}}(2x)^{-\frac{2}{3}}}{3} \bigg) - 36\sqrt[3]{2x}\sqrt{x+5}\ (4x-1)^8}{(4x-1)^9}$ 
$= \frac{\bigg(\frac{\sqrt[3]{2x}}{2\sqrt{x+5}} + \frac{\sqrt{x+5}}{6\sqrt[3]{2x}} \bigg) - 36\sqrt[3]{2x}\sqrt{x+5}\ (4x-1)^8}{(4x-1)^9}$ 
Suffice to say this is a mess, does a less-tedious way exist to find the derivative?

The classical trick is to consider
\begin{align}
\ln y = \frac{1}{3}\ln(2x)+\frac{1}{2}\ln(x+5)-9\ln(4x-1)
\end{align}
then differentiate to get
\begin{align}
\frac{y'}{y} =&\ \frac{1}{3x}+\frac{1}{2(x+5)}-\frac{36}{4x-1}\\
=&\ -\frac{196x^2+1045x+10}{6x(x+5)(4x-1)}.
\end{align}
Hence it follows
\begin{align}
y' =&\ -y\frac{196x^2+1045x+10}{6x(x+5)(4x-1)} = - \frac{\sqrt[3]{2x}\sqrt{x+5}}{(4x-1)^9}\frac{196x^2+1045x+10}{6x(x+5)(4x-1)} \\
=&\ - \frac{196x^2+1045x+10}{3\cdot 2^{2/3} x^{2/3}\sqrt{x+5}(4x-1)^{10}}
\end{align}
Remark: In the grand scheme of things, it doesn't save much time in terms of getting a simplified expression. However, it does have some organizational benefits. 
Remark $2$: As mentioned by egreg, we can also write
\begin{align}
\ln |y| = \ln\left|\frac{\sqrt[3]{2x}\sqrt{x+5}}{(4x-1)^9}\right| = \frac{1}{3}\ln|2x|+\frac{1}{2}\ln|x+5|-9\ln|4x-1|
\end{align}
and use the fact that 
\begin{align}
\frac{d}{dx}\ln|f(x)| = \frac{f'(x)}{f(x)}.
\end{align}
