$\displaystyle

y=\frac{v^3-2v\sqrt{v}}{v}

$

i've tried various techniques and keep getting something like..

$\displaystyle

=\frac{(3v^2-3v\sqrt{v})-(v^3-2v\sqrt{v})}{v^2}

$

im not sure how to get the answer, but BoB says the answer is

$\displaystyle

y'=\frac{2v-1}{\sqrt{v}}

$

help? maybe you can tell what im doing wrong or what i still need to do