Differentiation - Finding derivatives of this function - 
$h(y)= (y^{-3}+2y)(2+3y^3)^{-1/2} $ 
I tried and weren't sure on how to simplify it to get the answer 
$\frac{8-27y^{-1} - 12y^4-6y^3}{2(2+3y^3)^{3/2}} $ 
This is my try - 
$ (y^{-3} + 2y) (\frac{-1}{2} (2+3y^3)^{-3/2} (0+9y^2)) + (2+3y^3)(-3y^{-4} + 2) $ 
I simplified it till - 
$ (2+3y^3)^{-3/2}(-9/2 (y^{-3} + 2y) + (2+3y^3)^{5/2}(-3y^{-4} +2) $
$ \frac{(-9/2 (y^{-3} + 2y) + (2+3y^3)^{5/2}(-3y^{-4} +2)}{ (2+3y^3)^{3/2}}$
I'm not too sure how to simplify this further .. thanks ! 

There are two  typos in your calculation

We obtain applying the product rule
  \begin{align*}
&(y^{-3}+2y)(2+3y^3)^{-1/2}\\
&\qquad= (y^{-3} + 2y) \left(-\frac{1}{2} (2+3y^3)^{-3/2} (0+9y^2)\right) + (2+3y^3)^{\color{blue}{-1/2}}(-3y^{-4} + 2)\tag{1}\\
&\qquad= (2+3y^3)^{-3/2} \left(-\frac{9}{2}\color{blue}{y^2} (y^{-3} + 2y)+ {{(2+3y^3)}}(-3y^{-4} + 2)\right)\tag{2}\\
&\qquad=(2+3y^3)^{-3/2}\left(-\frac{9}{2}y^{-1}-9y^{3}-6y^{-4}-9y^{-1}+4+6y^3\right)\\
&\qquad=2(2+3y^3)^{-3/2}\left(8-27y^{-1} - 12y^4-6y^3\right)
\end{align*}
  and the claim follows.

Comment:


*

*In (1) a power $-\frac{1}{2}$ in $(2+3y^3)^{\color{blue}{-1/2}}$ was missing

*In (2) a factor $\color{blue}{y^2}$ was missing
