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Quick Question

  • Thread starter Altami
  • Start date
  • #1
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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 ?
 

Answers and Replies

  • #2
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would it look like....3v^2 - v^-1/2 ?
The 3v^2 part is correct! But how did you obtain v^-1/2??
 
  • #3
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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?
 
  • #4
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Well, [tex]\sqrt{x}=x^{1/2}[/tex], so you just need to apply the power rule!

But the problem is that you don't only have a square root, you have [tex]v\sqrt{v}[/tex]. To derive this, you have to apply the chain rule. (or if you know your algebra, you could notice that [tex]v^{3/2}=v\sqrt{v}[/tex] and immediately apply the power rule)...
 
  • #5
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Well, [tex]\sqrt{x}=x^{1/2}[/tex], so you just need to apply the power rule!

But the problem is that you don't only have a square root, you have [tex]v\sqrt{v}[/tex]. To derive this, you have to apply the chain rule. (or if you know your algebra, you could notice that [tex]v^{3/2}=v\sqrt{v}[/tex] 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.
 
  • #6
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3,277
Can't you just cancel v from the numerator and denominator??
 
  • #7
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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...
 
  • #8
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so if someone could show me? maybe a step by step...cause I have wasted a while on this one problem.....
 
  • #9
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So, nobody has an idea?
 
  • #10
33,075
4,776
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.

[tex]y = \frac{v^3 - 2v\sqrt{v}}{v} = v^2 - 2\sqrt{v} = v^2 - 2v^{1/2}[/tex]
 
  • #11
17
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No, you don't have to use the quotient rule if you simplify this first.

[tex]y = \frac{v^3 - 2v\sqrt{v}}{v} = v^2 - 2\sqrt{v} = v^2 - 2v^{1/2}[/tex]
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!
 
  • #12
33,075
4,776
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 - 2v3/2.
 
  • #13
17
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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 - 2v3/2.
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.
 
  • #14
33,075
4,776
Sure, you're welcome.
 

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