<s>[INST] 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!
 [/INST] While using the quotient rule for $f(x)=\dfrac{h(x)}{g(x)}\rightarrow f'(x)=\dfrac{h'(x)g(x)-h(x)g'(x)}{[g(x)]^2}$ is straightforward and gives
$y'=\dfrac{(2x+8)\sqrt{x}-\frac{1}{2}(x^2+8x+3)x^{-\frac{1}{2}}}{x}=(2x+8)x^{-\frac{1}{2}}-\frac{1}{2}(x^2+8x+3)x^{-\frac{3}{2}}=x^{-\frac{3}{2}}\left[(2x+8)x-\frac{1}{2}(x^2+8x+3)\right]=\dfrac{3x^2+8x-3}{2x^{-\frac{3}{2}}}$
we can also, with a simple manipulation, avoid the quotient rule:
$y=x^{\frac{3}{2}}+8\sqrt{x}+3x^{-\frac{1}{2}}$
$y'=\frac{3}{2}x^{\frac{1}{2}}+4x^{-\frac{1}{2}}-\frac{3}{2}x^{-\frac{3}{2}}=\frac{3x^2+8x-3}{2x^{-\frac{3}{2}}}$
</s>