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D/dx first principles

  1. Sep 19, 2006 #1
    I always wondered how you can differentiate a^x from first principles with the limit as dx approaches zero but I never managed to simplify it far enough to seperate dx on a different term. Can anyone help?
     
  2. jcsd
  3. Sep 19, 2006 #2

    radou

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    I guess it should look like that:

    [tex]
    \begin{equation*}
    \begin{split}
    &f(x) = a^x \\
    f'(x) &= lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}=lim_{h\rightarrow 0}\frac{a^{(x+h)}-a^x}{h} = \\
    &= lim_{h\rightarrow 0}\frac{a^xa^h-a^x}{h} = lim_{h\rightarrow 0}\frac{a^x(a^h-1)}{h} = \\
    &= a^x lim_{h\rightarrow 0}\frac{a^h-1}{h} = a^x lna \\
    \end{split}
    \end{equation*}
    [/tex]
     
    Last edited: Sep 20, 2006
  4. Sep 19, 2006 #3

    Hurkyl

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    You can always differentiate it using all of the tools you know, and then translate those tools into epsilon-deltas.

    radou: try something like:

    \begin{equation*}
    \begin{split}
    f(x) &= a^x \\
    f'(x) &= ... \\
    &= ... \\
    &= ...
    \end{split}
    \end{equation*}
     
  5. Sep 20, 2006 #4

    radou

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    Thanks. :smile:
     
  6. Sep 20, 2006 #5
    radou, you might want to break that process into seperate lines... I can't see half of it! :)
     
  7. Sep 20, 2006 #6

    HallsofIvy

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    Of course, the hard part is showing that
    [tex]\lim_{h\rightarrow 0}\frac{a^h-1}{h}= ln(a)[/tex]

    Many modern texts start by defining
    [tex]ln x= \int_1^x\frac{1}{t}dt[/tex]
    showing that this has all the properties of a natural logarithm, has an inverse, and then defining ex as its inverse.
     
  8. Sep 20, 2006 #7

    radou

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    Exactly. That's why I didn't prove it. :biggrin:
     
  9. Sep 20, 2006 #8
    Hey but that's not fair, any up to standard student can get there on his/her own! How do you actually solve the hard part? I think it's interesting for all those starting calculus...

    Tell me this at least... Can you diffentiate *every* equation from first principles? Even implicite ones?
     
  10. Sep 20, 2006 #9

    arildno

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    The simplest way to prove this rigourously is by introducing the ugly-looking function Exp(x):
    [tex]Exp(x)=1+\sum_{n=1}^{\infty}\frac{x^{n}}{n!}[/tex]
    where integral powers of numbers have been defined inductively.
    Exp(x) can be shown to have all the properties we would like an exponential function to have, including an inverse we call Log(x).
    Furthermore, we can differentiate Exp(x) termwise, yielding..Exp(x) itself.

    We therefore DEFINE [itex]a^{x}=Exp(x*Log(a))[/itex]
    and we may differentiate this by the use of the chain rule.
     
  11. Aug 26, 2010 #10

    Philip Wood

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    If we differentiate log x (to any base) from first principles, after a few lines of algebra we find it to be (1/x) times the log of the limit as h approaches zero of (1 + h)^(1/h). But this limit is the familiar definition of e. Job done. Using this result its then easy to find (using chain rule or whatever) the derivative of the inverse function, i.e. the exponential function.
    I'd say this counts as differentiating the exponential function from first principles. The limit Radou uses, is in my opinion, less well known that the limit I've referred to above.
     
  12. Aug 26, 2010 #11

    uart

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    Hi Philip. I'm pretty sure that lim h->0 (1 + h)^(1/h) = e is pretty much the same thing as showing that lim h->0 (e^h - 1)/h = 1. Think about it, one follows pretty easily from the other.
     
  13. Aug 26, 2010 #12

    Philip Wood

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    uart: Thank you. I agree that one follows from the other. Radou's limit, as he's tackling the general case of differentiating a^x, is of (a^h - 1)h, and comes to ln a; but this can also easily be shown to follow from lim h -> 0 (1 + h)^1/h. My point, though, is that lim h -> 0 (1 + h)^1/h arises naturally in the first principles differentiation of log x (base a) and is, arguably, the standard definition of e.
     
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