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Homework Help: Proof of derivative

  1. Dec 31, 2007 #1

    I was wondering if anyone could provide a proof for the following derivative

    [tex] f(x)=x^b [/tex]
    [tex] f^1(x)=bx^{b-1} [/tex]
    Last edited: Dec 31, 2007
  2. jcsd
  3. Dec 31, 2007 #2


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  4. Dec 31, 2007 #3
    erm the latex input isnt quite working out right in my first post sry
  5. Dec 31, 2007 #4
  6. Dec 31, 2007 #5


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    One can use Newton's generalized binomial theorem, which is covered on the wikipedia page, for that case.
  7. Dec 31, 2007 #6
    Alternatively, you can use a couple of cases

    Once you prove it for nonnegative integers, you can do this

    For negative integers:
    let [tex]n[/tex] be a positive integer

    [tex]x^{-n} = \frac{1}{x^n}[/tex]

    taking the derivative of x^-n is the same as taking the derivative on the right via the quotient rule

    [tex]\frac{d}{dx}\frac{1}{x^n}\right) = \frac{x^n(0) - 1(nx^{n-1})}{x^{2n}} = -nx^{-n-1}[/tex]

    proving it

    Now you can prove it for rationals through this:

    [tex]\frac{d}{dx} x^{p/q}[/tex]

    let [tex]y = x^{p/q}[/tex]

    [tex]y^q = x^p[/tex]

    implicitly differentiating

    [tex]qy^{q-1} \frac{dy}{dx} = px^{p-1}[/tex]

    [tex]\frac{dy}{dx} = \frac{px^{p-1}}{qy^{q-1}} = \frac{px^{p-1}}{qx^{p(q-1)/q}}[/tex]

    [tex]= \frac{p}{q} x^{p/q - 1}[/tex]

    can you figure out a similar method for irrationals too?
  8. Jan 1, 2008 #7
    I don't think there's a way to prove it for irrational exponents using the binomial theorem. Something else must be used.
  9. Jan 1, 2008 #8


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    But once you know it for rational exponents, the proof for irrational exponents is typical of arguments in anaysis.
  10. Jan 1, 2008 #9
    That is the "something else". You cannot prove it directly with the binomial theorem.
  11. Jan 1, 2008 #10

    Gib Z

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    Well to have used the binomial theorem to prove for rationals, you must have used the generalized version anyway, which applies to irrational exponents as well. You can prove it for rational exponents even without the binomial theorem, and after this its a simple matter of taking some limits.
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