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Question about calculus, on derivatives

  1. Jun 6, 2014 #1
    I am posting this question, in order to make something clear, since i am confused by derivation of the exponential function. I'll post the formula i used, correct me if you find something wrong thank you:
    [itex]
    {\frac{d}{Dx}}\ e^x =>
    [/itex]

    [itex]
    {\frac{e^{x + Dx} - e^x}{Dx}}\ =>
    [/itex]

    Here i factored out [itex] e^x [/itex]

    [itex]
    e^x\ {\frac{e^{Dx} - 1}{Dx}}\ =>
    [/itex]

    I integrated the exponential

    [itex] e^{Dx}\ =\ cosh(Dx)\ +\ sinh(Dx)\ [/itex]

    Now based on theory
    [itex] e^{Dx} [/itex]
    already approaches a value which is a tiny little above 1 so if we subtract one it will be one tiny little value above 0 and if we divide by [itex] Dx [/itex]
    we are likely to get 1 as the values approach each other as they shrink to lower values.
    But as i try to factor for [itex] sinh(x) [/itex] i take that since
    [itex]
    coth(x)\ =\ {\frac{\cosh(x)}{sinh(x)}}\
    [/itex]
    i take that
    [itex] cosh(Dx)\ =\ coth(Dx)\ *\ sinh(Dx) [/itex]

    [itex]
    e^x\ {\frac{coth(Dx)sinh(Dx) + sinh(Dx) - 1}{Dx}}\ =>
    [/itex]

    [itex]
    e^x\ {\frac{sinh(Dx)\ (coth(Dx) + 1)}{Dx}}\ -\ {\frac{1}{Dx}}\
    [/itex]

    As sinh(Dx) approaches 0 so does Dx and they tend to be 1 in division
    [itex]
    e^x\ *\ (coth(Dx)\ +\ 1)\ -\ {\frac{1}{Dx}}\
    [/itex]
     
    Last edited: Jun 6, 2014
  2. jcsd
  3. Jun 6, 2014 #2
    I was gonna try to help, but every time I refreshed, your post had changed. I'm not going to try to hit a moving target.
     
  4. Jun 6, 2014 #3

    verty

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    This is too sloppy, there are brackets missing, ##D## should be ##Δ##, there are some questionable manipulations and I don't know why you are going in this direction.
     
  5. Jun 6, 2014 #4

    SteamKing

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    This is how to write the derivative correctly:

    This next step is incorrect:

    [tex]{\frac{e^{x + Dx} - e^x}{Dx}}\ =>[/tex]

    The derivative is defined as

    [tex]
    {\frac{d}{dx}}\ e^x =
    \lim_{Δx\rightarrow 0}{\frac{e^{x + Δx} - e^x}{Δx}}[/tex]

    You can apply the Law of Exponents here to factor out [itex]e^{x}[/itex], leaving

    [tex]
    {\frac{d}{dx}}\ e^x = e^x
    \lim_{Δx\rightarrow 0}{\frac{e^{Δx} - 1}{Δx}}[/tex]

    When you evaluate this limit, it becomes an indeterminate form, [itex]{\frac{0}{0}}[/itex], which can be treated with L'Hopital's Rule.

    I have no idea how you got to the next step. In general, finding derivatives does not involve the use of integration:

     
  6. Jun 6, 2014 #5

    lurflurf

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    ^L'Hopital's Rule is not needed
    We need to know the derivative of e^x when x is 0.
    That is a whole other proof.
    Then

    $$\exp^\prime(0)=\lim_{h\rightarrow 0}\frac{\exp(0+h)-\exp(0)}{h}\\
    \exp^\prime(0)\exp(x)=\lim_{h\rightarrow 0}\frac{\exp(0+h)-\exp(0)}{h}\exp(x)\\
    \exp^\prime(0)\exp(x)=\lim_{h\rightarrow 0}\frac{\exp(x+h)-\exp(x)}{h}\\
    \exp^\prime(0)\exp(x)=\exp^\prime(x)
    $$
     
  7. Jun 6, 2014 #6

    SteamKing

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    I see what you are trying to do, but I think you are boot-strapping.

    You haven't rigorously shown that the following limit exists:

    [tex]\lim_{Δx\rightarrow 0}{\frac{e^{Δx} - 1}{Δx}}[/tex]

    In studying derivatives, one doesn't normally start off with trying to show that e^x is its own derivative. I still think L'Hopital's Rule is one call you can make here to complete the proof.
     
  8. Jun 6, 2014 #7

    lurflurf

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    ^That is my point I assume exp'(0), you assume exp'(x).
    e^x is its own derivative, what else would you show?
    L'Hopital's rule is a strong result that the OP may or may not know, using it also requires knowing exp'(x).
    In any case we must know exp'(0).

    To show exp'(0) we need to know what definition of exp(x) the OP is using.
     
  9. Jun 7, 2014 #8

    HallsofIvy

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    One can show, as SteamKing did, but using an arbitrary positive number, a, rather than e, that
    [tex]\frac{da^x}{dx}= \left(\lim_{\Delta x\to 0} \frac{a^{\Delta x}- 1}{\Delta x}\right)a^x[/tex]
    and that the limit exists, for any positive a, can be shown. That is, that the derivative of [itex]a^x[/itex] is a constant (depending on a) times [itex]a^x[/itex]. One can show that the limit is less than 1 for, say a= 2, larger than 1 for a= 3, and depends continuously on a. That means that there exist a value of a, between 2 and 3, such that the limit is 1 and we define "e" to be that value.

    But many modern Calculus texts do the derivatives in the opposite way- define the natural logarithm function as [itex]\ln(x)= \int_0^1 \frac{dt}{t}[/itex], show that the inverse function to this new function is some number to the x power and then define "e" to be that number.
     
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