# Differentiate a function using the quotient rule

1. Jul 20, 2012

### frosty8688

1. Differentiate

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

3. I am trying to use the quotient rule, but am having trouble understanding how to use the square root. Here's what I have $\frac{v(3v^{2}-2*1/2\sqrt{v}) - v^{3}+2v*1/2\sqrt{v}}{v^{2}}$

2. Jul 20, 2012

### Mark M

Remember that the square root of a number is equivalent to that number to the one-half power. So, $$\sqrt {x} = x^{\frac {1} {2}}$$ Replace the square roots with exponents, simplify the numerator and then use the power rule.

3. Jul 20, 2012

### eumyang

$\sqrt{v} = v^{1/2}$, so
$2v\sqrt{v} = 2v\;v^{1/2} = ?$

EDIT: Beaten to it.

4. Jul 20, 2012

### frosty8688

I did, that's how I got $\sqrt{v}$ to equal $\frac{1}{2\sqrt{v}}$. I am just having problems figuring out how to multiply it out.

5. Jul 20, 2012

### frosty8688

It equals v*v$^{-1/2}$

6. Jul 20, 2012

### eumyang

Before you start differentiating, rewrite the expression without square roots:
$$y = \frac{v^3 - 2v\;v^{1/2}}{v}$$
So what is
$2v\;v^{1/2}$?

7. Jul 20, 2012

### Robert1986

Is there a reason you are using the quotient rule?

8. Jul 20, 2012

### eumyang

If you were answering me, then I'm afraid you're wrong. Use the properties of exponents.

9. Jul 20, 2012

### eumyang

I was wondering the same thing, but I assumed that this is a problem from a textbook and the directions required the use of the quotient rule.

10. Jul 20, 2012

### Robert1986

That's what I was leaning toward - but I just want to be sure.

11. Jul 20, 2012

### frosty8688

It would be 2*$\frac{3}{2}$v$^{1/2}$ which would equal 3v$^{1/2}$

12. Jul 20, 2012

### eumyang

Well, that's the derivative of $2v^{3/2}$, but I was expecting you to say $2v^{3/2}$ first, before taking the derivative. So now the original expression becomes
$$y = \frac{v^3 - 2v^{3/2}}{v}$$
Use the quotient rule from here.

13. Jul 20, 2012

### frosty8688

I see. Thanks for helping.

14. Jul 20, 2012

### Mark M

I don't believe you calculated the derivative of the numerator correctly

$v^{3} - 2v \sqrt {v}$
$v^{3} - 2v v^{\frac {1} {2}}$
$v^{3} - 2v^{\frac {3} {2}}$
Using the sum rule, differentiate both to get
$3v^{2} - 3 \sqrt {v}$

Use the quotient rule as you normally did, see if it's easier to simplify.

EDIT: You beat me to it this time, you got your revenge.

15. Jul 20, 2012

### frosty8688

Can't you just simplify the equation and differentiate that?

16. Jul 20, 2012

### eumyang

On another forum (that also provided help in math), I recall that you could see who was viewing a particular thread. I remember that I would not bother replying to a thread if I saw that someone else was viewing the same thread.

What do you mean? Are you now saying that we don't have to use the quotient rule?

17. Jul 20, 2012

### frosty8688

I'm just saying that wouldn't it be simpler to divide the numerator by v and differentiating the equation after that. But in the book it expects us to use the quotient rule.

18. Jul 20, 2012

### eumyang

Of course it would be simpler. I guess you haven't read all of the posts in this thread:

19. Jul 20, 2012

### frosty8688

Here's what I got using the quotient rule $\frac{2v^{3}-v^{3/2}}{v^{2}}$ which equals 2v -$\frac{1}{\sqrt{v}}$

20. Jul 20, 2012

### eumyang

Correct.

BTW, there's nothing wrong with putting an entire expression inside the tex tags. So instead of writing
2v -$\frac{1}{\sqrt{v}}$
you could write
$2v - \frac{1}{\sqrt{v}}$
Looks better, doesn't it?