Differentiating (v^3-2v*squareroot v): A Try

[Altami]

Homework Statement

How would you differentiate, (v^3-2v*squareroot v)?

Homework Equations

The Attempt at a Solution

would it look like...3v^2 - v^-1/2 ?

 

[micromass]

would it look like...3v^2 - v^-1/2 ?

The 3v^2 part is correct! But how did you obtain v^-1/2??

 

[Altami]

The 3v^2 part is correct! But how did you obtain v^-1/2??

Ah, see that's the thing how to you differentiate a square root? I have no idea?

 

[micromass]

Well, \sqrt{x}=x^{1/2}, so you just need to apply the power rule!

But the problem is that you don't only have a square root, you have v\sqrt{v}. To derive this, you have to apply the chain rule. (or if you know your algebra, you could notice that v^{3/2}=v\sqrt{v} and immediately apply the power rule)...

 

[Altami]

Well, \sqrt{x}=x^{1/2}, so you just need to apply the power rule!

But the problem is that you don't only have a square root, you have v\sqrt{v}. To derive this, you have to apply the chain rule. (or if you know your algebra, you could notice that v^{3/2}=v\sqrt{v} and immediately apply the power rule)...

Well okay, I wanted to see if by understanding how to derive squareroot v, I could understand a much bigger problem.

The actual problem that is connected is this

y=(v^3 - 2v*square rootv)/ v

and I have to differentiate it, I know I have to use the quotient rule and I have tired but the dang square root is cause me problems.

 

[micromass]

Can't you just cancel v from the numerator and denominator??

 

[Altami]

Can't you just cancel v from the numerator and denominator??

I don't think so, I haven't tried that I know the answer...because it is...(2v-1)/squareroot2

But I don't know how to get to that, I've tried everything...that I know of...and I still can't get that answer...

 

[Altami]

so if someone could show me? maybe a step by step...cause I have wasted a while on this one problem...

 

[Altami]

So, nobody has an idea?

 

[Mark44]

y=(v^3 - 2v*square rootv)/ v

and I have to differentiate it, I know I have to use the quotient rule and I have tired but the dang square root is cause me problems.

No, you don't have to use the quotient rule if you simplify this first.

y = \frac{v^3 - 2v\sqrt{v}}{v} = v^2 - 2\sqrt{v} = v^2 - 2v^{1/2}

 

[Altami]

No, you don't have to use the quotient rule if you simplify this first.

y = \frac{v^3 - 2v\sqrt{v}}{v} = v^2 - 2\sqrt{v} = v^2 - 2v^{1/2}

How can I vote you to be a life saver? Thank you so much, for some reason my teacher told me I HAD to use the quotient rule and I guess that blocked me to thinking of other ways! Thank you so much, your a life saver!

 

[Mark44]

Well, if the instructions are that you have to use the quotient rule, then I haven't been any help at all. If so, I would write the numerator as v³ - 2v^3/2.

 

[Altami]

Well, if the instructions are that you have to use the quotient rule, then I haven't been any help at all. If so, I would write the numerator as v³ - 2v^3/2.

Well, I meant that he didn't really give us clues as to have other ways of solving them, your were a great help. I still had to differentiate the final answer which you gave me, but I was able to do that no problem. Thank you anyways, again you did help me.

 

[Mark44]

Sure, you're welcome.