derivative with square root I have been trying to figure this equation for some time now, but have come up empty. I have tried multiple ways on solving it. Whether by using the Quotient Rule or some other method, I can't seem to figure it out. Any help would be appreciated.
Find the derivative of the function 
$
y = \frac{x^{2} + 8x + 3 }{\sqrt{x}}
$
The answer is:
$
y' = \frac{3x^{2}+ 8x - 3 }{2x^\frac{3}{2}}
$
I'm having trouble getting to that answer. So if someone could point me in the right direction by showing the steps, I would be grateful. Thanks!

For differentiating $\sqrt x$ one should rewrite it as $x^{\frac{1}{2}}$ and use the power rule. Given $y=\frac{x^2+8x+3}{\sqrt x}$ we have:
$$y'=\frac{\left(\sqrt x\right)\left(2x+8\right)-\left(\frac{1}{2}x^{-\frac{1}{2}}\right)\left(x^2+8x+3\right)}{\left(\sqrt x\right)^2}
\\=x^{-\frac{3}{2}}\left(2x^2+8x-\frac{1}{2}x^2-4x-\frac{3}{2}\right)
\\=\frac{1}{2}x^{-\frac{3}{2}}\left(3x^2+8x-3\right)$$
