Differentiate the following function: $y = \frac{2x^2 + 6\sqrt{x} }{9x}$ My answer is $- \dfrac{1}{9 \sqrt{x} }$, however, the program I am using states that I am wrong. Where have I went wrong? 
 A: $y=2x^2+\frac{2}{3}x^{-1/2}$. Now use the power rule to differentiate.
A: $4x-\frac{2}{3}\cdot \frac{1}{2}x^{-\frac{3}{2}}$
A: $$y = 2x^2 + \frac{6\sqrt x}{9x} = 2x^2 +\frac {2}{3\sqrt x} = 2x^2 + \frac 23\cdot x^{-1/2}$$
$$y' = 4x - \frac 12\cdot\frac 23 \cdot x^{-3/2} =  4x - \frac{1}{3x^{3/2}}$$
EDIT: 
Given the OP's comments and attempted edit, it seems that the function of interest is intended to be $$y = \frac{2x^2 + 6\sqrt{x}}{9x}= \frac 29 x + \frac 23\cdot x^{-1/2}$$
If so, then
$$\begin{align} \frac{dx}{dy} & = \frac 29 - \frac 12\cdot\frac 23 \cdot x^{-3/2}\\ \\ 
& =  \frac 29 - \frac{1}{3x^{3/2}}\\ \\
& = \frac{2x^{3/2} - 3}{9x^{3/2}}
\end{align}$$
A: The derivative of your first term is just $4x$ and if you use the power rule for the second one, you get $-\frac{1}{3}\times x^{-\frac{3}{2}}$ so eventually you get 
$$4x-\frac{1}{3}\times x^{-\frac{3}{2}}$$
A: $y = \frac{2x^{2}}{9x} + \frac{6x^\frac{1}{2}}{9x}=\frac{2x}{9} + \frac{2}{3}x^{-\frac{1}{2}}$.
Then $\frac{dy}{dx}=\frac{2}{9} -\frac{2}{3}⋅-\frac{1}{2}x^{-\frac{3}{2}}$
