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I'm stuck here derivative.

  1. Oct 20, 2004 #1

    JasonRox

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    x^(1/3)

    The only way I have for this on this is to show that the derivative of x^n equals Nx^(n-1).

    I can't think of a way to do it in first principles.

    If I can just break up (x+h)^1/3, I'm good.

    If you break that up, you get...

    [tex]x^\frac{1}{3} + 1/3x^\frac{2}{9}h^\frac{1}{9} + 1/3x^\frac{1}{9}h^ \frac{2}{9}+ h^\frac{1}{3}[/tex]

    It's probably wrong, but that is all I can think of.

    Any pointers would help.
     
  2. jcsd
  3. Oct 20, 2004 #2
    You're just trying to get the derviative of [itex]x^{1/3}[/itex]

    You have the procedure written down correctly in the second line then you went way overboard after that.

    [tex]\frac{d}{dx}x^n=nx^{n-1}[/tex]
    where n is a constant

    Replace n with 1/3--that's all there is to it.

    Is your question written correctly?
     
  4. Oct 20, 2004 #3

    Fredrik

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    I don't think he's trying to calculate the derivative. He's trying to prove that the rule works even when n is not an integer (specifically when n=1/3).

    Jason, are you allowed to use what you know about how to take the deriviate of exponentials, logarithms and composite functions? If you are, I suggest that you use

    [tex]x^{\frac{1}{3}}=e^{\frac{1}{3}\ln x}[/tex]
     
    Last edited: Oct 20, 2004
  5. Oct 20, 2004 #4

    JasonRox

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    Actually know we can not do that.

    This must be used doing first principles. I know how to find derivatives.

    We must use h approaches 0 or x approaches a.

    That is why I wrote (x+h)^1/3, which is from the h approaches 0 method.

    Obviously we have (x+h)^1/3-x^1/3 for the numerator. That is why I broke it up in the first post, so I can get rid of the h in the numerator, but since I can't do that because (h^1/3)/h = 1/h^2/3, according to exponent laws. This is where I am stuck.

    If I had an exam right now, I would have to do the x^n -> Nx^(n-1) method, but that is not practical nor what they are asking for.

    Any tips are appreciated.

    Note: We must assume we don't "know" the derivative of ln x in this case.
     
  6. Oct 20, 2004 #5

    JasonRox

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    So I guess I'm not the only who is stuck.
     
  7. Oct 20, 2004 #6

    arildno

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    Dearly Missed

    I remember a thread here at PF recently:
    Set:
    [tex]a=(x+h)^{\frac{1}{3}},b=x^{\frac{1}{3}}[/tex]
    Then, for the derivative:
    [tex]\frac{a-b}{h}=\frac{(a-b)(a^{2}+ab+b^{2})}{(a^{2}+ab+b^{2})h}=[/tex]
    [tex]\frac{a^{3}-b^{3}}{(a^{2}+ab+b^{2})h}=\frac{h}{(a^{2}+ab+b^{2})h}=[/tex]
    [tex]\frac{1}{(a^{2}+ab+b^{2})}\to\frac{1}{3x^{\frac{2}{3}}} (h\to{0})[/tex]
     
  8. Oct 20, 2004 #7

    JasonRox

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    You're my hero! :blushing:

    Thanks!

    Gave me a whole new perspective on how to do this.
     
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