Differentiate a function using the quotient rule

[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 $\frac{v(3v^{2}-2*1/2\sqrt{v}) - v^{3}+2v*1/2\sqrt{v}}{v^{2}}$

 

[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.

 

[eumyang]

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

$\sqrt{v} = v^{1/2}$, so
$2v\sqrt{v} = 2v\;v^{1/2} = ?$EDIT: Beaten to it.

 

[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.

 

[frosty8688]

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

 

[eumyang]

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

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

 

[Robert1986]

Is there a reason you are using the quotient rule?

 

[eumyang]

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

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

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

 

[eumyang]

Is there a reason you are using the quotient rule?

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.

 

[Robert1986]

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.

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

 

[frosty8688]

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

 

[eumyang]

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

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.

 

[frosty8688]

I see. Thanks for helping.

 

[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.

 

[frosty8688]

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

 

[eumyang]

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.

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.

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

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

 

[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.

 

[eumyang]

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.

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

Is there a reason you are using the quotient rule?

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.

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

 

[frosty8688]

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

 

[eumyang]

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

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?