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Derivative of Exponential Function

  1. Jul 20, 2012 #1
    I am reading the about the derivative of an exponential function using the limit definition, but one step I don't quite understand: [itex]lim_{h\rightarrow0}\frac{a^h -1}{h} = f'(0)[/itex] Wouldn't that limit equal 0/0?
     
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  3. Jul 20, 2012 #2

    HallsofIvy

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    Every derivative limit is of the form "0/0" but, of course, the limit itself is NOT "0/0".
     
  4. Jul 20, 2012 #3
    Oh, yes, how could I be so foolish to forget. Well, what sort of algebraic manipulation am I to perform then?
     
  5. Jul 20, 2012 #4
    Have you defined the exponential function as [itex]\sum\frac{z^n}{n!}[/itex]. If so, use that to prove the limit.
     
  6. Jul 21, 2012 #5

    HallsofIvy

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    There's the rub! It depends, of course, upon how you define [itex]e^x[/itex].

    Many modern Calculus texts first define [itex]ln(x)= \int_a^x dx/x[/itex] and then define [itex]e^x[/itex] as the inverse function so your question never arises. That's my preferred method.

    Robert1986 suggests that you define [itex]e^x[/itex] as as power series. That's perfectly reasonable, and again, your question does not arise.

    But it is probably most common to define [itex]a^x[/itex] by first defining [itex]a^n[/itex] by "a multiplied by itself n times" then extending to all real numbers by requiring that [itex]a^{n+m}= (a^n)(a^m)[/itex], [itex](a^n)^m= a^{mn}[/itex], and continuity. Using that definition, we have
    [tex]\frac{f(x+h)-f(x)}{h}= \frac{a^{x+h}- a^x}{h}= \frac{a^xa^h- a^x}{h}[/tex]
    [tex]= \frac{a^x(a^h- 1)}{h}= a^x\frac{a^h- 1}{h}[/tex]

    So that
    [tex]\frac{da^x}{dx}= \lim_{h\to 0}\frac{a^{h+1}- a^x}{h}= a^x\lim_{h\to 0}\frac{a^h- 1}{h}[/tex]
    is just [itex]C_aa^x[/itex] where [itex]C_a[/itex] is that limit.

    It is easy to show that if 0< a< 1, then [itex]C_a[/itex] is negative, that [itex]C_2[/itex] is less than 1, that [itex]C_3[/itex] is greater than 1, and that [itex]C_a[/itex] continuously increases as a increases. That means that there must exist some value of a, between 2 and 3, such that [itex]C_a= 1[/itex] and we define "e" to be that value of a.

    From [itex]\lim_{h\to 0}\frac{e^h- 1}{h}= 1[/itex] we can say that, for h close to 0, [itex]\frac{e^h- 1}{h}[/itex] is close to 1 so that [itex]e^h- 1[/itex] is close to h. Then [itex]e^h[/itex] is close to 1+ h and so e is close to [itex](1+ h)^{1/h}[/itex] which is common limit definition of e: [itex]e= \lim_{h\to 0}(1+ h)^{1/h}[/itex] or, taking n= 1/h, [itex]e= \lim_{n\to\infty} (1+ 1/n)^n[/itex].
     
    Last edited: Jul 21, 2012
  7. Jul 23, 2012 #6
    @HallsOfIvy:

    So this constant you are speaking about is the same as f'(0)?

    Edit:

    Also, the books says that this limit is the value of the derivative at zero. Why is that? And since the derivative of an exponential function itself, wouldn't that always be equal to one?
     
    Last edited: Jul 23, 2012
  8. Jul 24, 2012 #7
    It's not; f'(0) is the derivative of f(x) evaluated at x = 0. The constant Ca, in a way, turns out to be a "function" of a that changes the value of f'(0) depending on the value of a in f(x) = ax.

    If f(x) = ax, then the difference quotient for f(x) is[tex]\frac{a^{x+h} - a^x}{h} = a^x\frac{a^h - 1}{h}[/tex]
    At x = 0, the difference quotient turns out to be (ah - 1)/h, and taking the limit as h→0 gives the derivative at 0, written as f'(0).
     
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