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I am attempting to find the derivative of $\frac{(v^3 -2v\sqrt v) } v$ using the quotient rule I set the problem up like this: $v(3v^2 - 3v^{1/2}) - (v^3 - 2v^{3/2})1$ and the denominator is $v^2$ which I do not think it important until the end. Anyways from that I get $ \frac {2v^3 - 2v^{3/2}}{ v^2} $.

This is not the answer my book is getting, I am sure I messed up something simple.

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    After dividing each term of $2v^3 - v^{3/2}$ by $v^2$, you do get $2v - (1/\sqrt{v})$. By the way, your original post seems to have $\frac{2v^3 - 2v^{3/2}}{v^2}$, not the correct $\frac{2v^3 - v^{3/2}}{v^2}$.2011-09-20

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Check your simplification, because your first steps were correct. But your simplification is close (not quite). In other words, $3-2$ is very often less than $2$.

Seeing the answer your book gave, I wanted to also point out that there isn't much reason to use the quotient rule here. $\frac{(v^3 -2v\sqrt v) } v = v^2 - 2\sqrt v$, which is easier to differentiate.

Of course, the quotient rule has its place.

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    @Jordan: the second fraction is $\dfrac{v^{3/2}}{v^2}=v^{3/2-2}=v^{-1/2}$2011-09-20
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$\begin{align} \frac{\operatorname{d}\frac{v^3-2v\sqrt v}{v}}{\operatorname{d}v} &=\frac{v(3v^2-2\sqrt{v}-2\cdot\frac{v}{2\sqrt{v}})-(v^3-2v\sqrt{v})}{v^2}\\ &=\frac{3v^3-2v\sqrt{v}-\frac{v^2}{\sqrt{v}}-v^3+2v\sqrt{v}}{v^2}\\ &=\frac{v^2(2v-\frac{1}{\sqrt{v}})}{v^2}\\ &=2v-\frac{1}{\sqrt{v}} \end{align}$

You can also argue that $v>0$ since it is under a square root, and in the denominator, therefore we can reduce the original function:

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

From here it is very straight forward to derive and have the same result.