Finding derivatives using the quotient rule

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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$(v(3v^2 - 3v^{1/2}) - (v^3 - 2v^{3/2})1)/v^2$ is correct, but it is not equal to $\frac {2v^3 - 2v^{3/2}} v$. –  Américo Tavares Sep 20 '11 at 22:05
Then I will do this step here, I have done it many times and got the same answer every time. Likely I am making a simple math error because I am so bad at math. Anyways the numerator should come out to $3v^3 - 3v^{3/2} - v^3 + 2v^{3/2}$ which gives $2v^3 -v^{3/2}$ –  user138246 Sep 20 '11 at 22:10
Could you also post the answer from the book? –  Shaun Ault Sep 20 '11 at 22:13
I am still getting $-v^{2/3}$ –  user138246 Sep 20 '11 at 22:13
The book got $2v- ( 1/ \sqrt v)$ –  user138246 Sep 20 '11 at 22:14

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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That is what I was getting, must have mistyped it here. But the book is giving me a v^-1/2 and I am getting 3/2 –  user138246 Sep 20 '11 at 22:18
@Jordan: then you should divide by $v^2$, and you see that your answer and the book's answer are the same. –  mixedmath Sep 20 '11 at 22:18
I don't know how to do that. –  user138246 Sep 20 '11 at 22:19
@Jordan: $\frac {2v^3 - v^{3/2}}{ v^2} = \frac{ 2v^3}{v^2} - \frac{v^{3/2}}{v^2} = ...$ - and you can finish from there? –  mixedmath Sep 20 '11 at 22:20
$2v - v^1/2$ I think. –  user138246 Sep 20 '11 at 22:23

\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.

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