Derivation with x^2 and root x

  • #1
disregardthat
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Homework Statement



derivate: [tex]x^2 \sqrt{x}[/tex]

Homework Equations



[tex](u \cdot v)^\prime = u^\prime \cdot v + v^\prime \cdot u[/tex]

The Attempt at a Solution



I found:
[tex](x^2)^\prime = 2x[/tex]
[tex](\sqrt{x})^\prime = \frac{1}{2 \sqrt{x}}[/tex]

I entered them into the equation:

[tex]2x \cdot \sqrt{x} + x^2 \cdot \frac{1}{2 \sqrt{x}} = 2x \sqrt{x} + \frac{x^2}{2 \sqrt{x}}[/tex]

But this is not correct!
 

Answers and Replies

  • #2
Hurkyl
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Why do you think it is not correct?
 
  • #3
radou
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Try to get rid of the square root in the denominator.
 
  • #4
robphy
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Can you simplify your last expression?
 
  • #5
disregardthat
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Ok, sorry. I checked it on my calculator many times and never got the correct result. But this time I did, by adding some extra brackets.. My calc just won't accept things with having it stuffed in with brackets!

Simplify? I guess you could simplify it:
[tex]2x \sqrt{x} + \frac{x^2}{2 \sqrt{x}} = \frac{4x^{1.5}}{2} + \frac{x^{1.5}}{2} = \frac{5x^{1.5}}{2} = \frac{5x \sqrt{x}}{2}[/tex]

Jesus christ, now my textbook's answer is correct too!
Well, I guess that's a good thing. I must have seen on it uncorrectly. I got it right now... Much making a topic about ...
 
  • #6
Hurkyl
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FYI, you can do this problem without the product rule -- do you see how?
 
  • #7
disregardthat
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When you say it I think so, yes.

[tex]x^2 \sqrt{x} = x^{2+0.5} = x^{2.5} = x^{\frac{5}{2}}[/tex]

[tex](x^{\frac{5}{2}})^\prime = \frac{5}{2} \cdot x = \frac{5x}{2}[/tex]
 
  • #8
robphy
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When you say it I think so, yes.

[tex]x^2 \sqrt{x} = x^{2+0.5} = x^{2.5} = x^{\frac{5}{2}}[/tex]

[tex](x^{\frac{5}{2}})^\prime = \frac{5}{2} \cdot x = \frac{5x}{2}[/tex]
missing exponent?
 
  • #9
disregardthat
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Of course
[tex]\frac{5x^{\frac{3}{2}}}{2}[/tex]
 

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