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Derivation rule Dx^r=rx^(r-1)

  1. Jan 25, 2008 #1
    How do you prove the derivative

    [tex]
    Dx^r = rx^{r-1}
    [/tex]

    for [itex]r\in\mathbb{R}\backslash\mathbb{Q}[/itex]?
     
    Last edited: Jan 25, 2008
  2. jcsd
  3. Jan 25, 2008 #2
    I know how to prove

    [tex]
    f(x) = x^{n/m}\quad\implies\quad f'(x) = \frac{n}{m}x^{n/m\;-1}
    [/tex]

    for [itex]n,m\in\mathbb{Z}[/itex], but since the derivative mapping is not continuous, you cannot easily commute the limit and derivative like this

    [tex]
    D x^r = D(\lim_{q\to r} x^q}) \underset{?}{=} \lim_{q\to r}(Dx^q) = \lim_{q\to r} qx^{q-1} = rx^{r-1}
    [/tex]

    For rational exponents you can carry out the proof by taking the derivative of the both sides of

    [tex]
    (f(x))^m = x^n
    [/tex]

    and then solving [itex]f'(x)[/itex].
     
  4. Jan 25, 2008 #3

    gel

    User Avatar

    That's fine, just as long as you can show that the limit on the right hand side converges uniformly (on an open interval containing x).
     
  5. Jan 25, 2008 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    If you can differentiate exponentials and logarithms, then you can write [itex]x^r= e^{r ln(x)}[/itex]. By the chain rule, the derivative of that is [itex](r/x)e^{r ln(x)}= (r/x)x^r= r x^{r-1}[/itex]. Of course the derivatives of ex and ln(x) can be defined without reference to xr.
     
  6. Jan 25, 2008 #5
    Is this true:

    Let [itex]I\subset \mathbb{R}[/itex] be some open interval. Let [itex]f_n:I\to\mathbb{R}[/itex] be a sequence of differentiable functions, such that [itex]f_n\to f[/itex] uniformly, where [itex]f[/itex] is some function, and such that [itex]f'_n[/itex] converges uniformly towards something. Then [itex]f[/itex] is differentiable and [itex]f'_n\to f'[/itex].

    ?


    Surprising trick!
     
  7. Jan 25, 2008 #6

    gel

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    What HallsOfIvy said is probably the best way to approach this problem, but yes this is true. One way to prove it is to use the mean value theorem.

    [tex]
    f_n(x+h)=f_n(x) + h f_n^\prime(x+a_n)
    [/tex]

    where |a_n|<|h|. If [itex]f_n[/itex] converges uniformly to g and a is a limit point of a_n then

    [tex]
    f(x+h)=f(x) + h g(x+a)
    [/tex]

    Assuming g is continuous and using |a|<=|h| gives [itex]f^\prime(x)=g(x)[/itex].
     
    Last edited: Jan 25, 2008
  8. Jan 26, 2008 #7
    You mean "if [itex]f'_n[/itex] converges uniformyl to g"? But is there any reason to assume that [itex]\lim_{n\to\infty}a_n[/itex] exists?

     
  9. Jan 26, 2008 #8

    gel

    User Avatar

    Yes, that's what I should have said. You don't need to assume that a_n converges, as it is always possible to pass to a subsequence such that this is true. Alternatively use

    [tex]
    \inf_{|a|<|h|}f_n^\prime(x+a)\le(f_n(x+h)-f_n(x))/h\le \sup_{|a|<|h|}f_n^\prime(x+a)
    [/tex]

    then take limits as n->infinity

    [tex]
    \inf_{|a|<|h|}g(x+a)\le(f(x+h)-f(x))/h\le \sup_{|a|<|h|}g(x+a)
    [/tex]

    and letting h->0 gives [itex]f^\prime(x)=g(x)[/itex].
     
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