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Why does d e^x / dx = e^x ?

  1. Nov 16, 2005 #1
    why does d e^x / dx = e^x ?
     
  2. jcsd
  3. Nov 16, 2005 #2

    mathwonk

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    the fact that e^(x+y) = e^x e^y, implies that, if e^x is differentiable, then the derivative is a constant times e^x.


    the explicit value of the constant depends on the base. the base e is defined as the unique base such that the constant equals 1.
     
  4. Nov 16, 2005 #3
    [tex]\frac{d}{dx}\,e^{x}=\lim_{h\to 0}\frac{e^{x}\left(e^{h}-1\right)}{h}[/tex]

    [tex]e=\lim_{h\to 0}\left(1+h\right)^{\frac{1}{h}}\implies e^{h}=\lim_{h\to 0}\left(1+h\right)[/tex]

    Substitute:

    [tex]\frac{d}{dx}\,e^{x}=\lim_{h\to 0}\frac{e^{x}\left(\left(1+h\right)-1\right)}{h}=\lim_{h\to 0}e^{x}=e^{x}[/tex]
     
  5. Nov 16, 2005 #4
    because e^a=e^a*a' and x'=1*dx/dx which is 1
     
  6. Nov 16, 2005 #5

    benorin

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    Is power series too much?
     
  7. Nov 16, 2005 #6
    Power rule doesn't work for variables.

    e^x=> xe^(x-1) is not right as far as i can tell.
     
  8. Nov 17, 2005 #7

    Integral

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    As Binorin said.
    [tex] e = 1 + x + \frac {x^2} {2!} + \frac {x^3} {3!} +\frac {x^4} {4!} + ... [/tex]

    Now differentiate.
     
  9. Nov 17, 2005 #8
    0 = e^x ? :surprised :wink:
     
  10. Nov 17, 2005 #9
    Power series != power rule.
     
  11. Nov 17, 2005 #10
    Hehe, its probably just a typo :biggrin: , it should be:
    [tex]e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots[/tex]

    (offcourse :)
     
  12. Nov 17, 2005 #11

    Integral

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    You may be correct :blushing:
     
  13. Nov 18, 2005 #12


    [tex]\frac{d}{dx} (e^x) = e^x[/tex] ???

    Can it not be stated that that the solution is derived from theorem?
    [tex]\frac{d}{dx} (a^x) = a^x \ln a[/tex]
    [tex]a = e[/tex]
    [tex]\frac{d}{dx} (e^x) = e^x \ln e[/tex]
    [tex]\ln e = 1[/tex]
    therefore:
    [tex]\boxed{\frac{d}{dx} (e^x) = e^x}[/tex]
     
    Last edited: Nov 18, 2005
  14. Nov 18, 2005 #13

    shmoe

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    Probably not as that theorem is usually proved by using the derivative of e^x.
     
  15. Nov 18, 2005 #14
    I think series subject is the better one
     
  16. Nov 18, 2005 #15

    D H

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    Why is [tex]\frac d {dx} \exp(x) = \exp(x)[/tex]? The answer depends on how one defines the exponential function.

    1. The exponential function is the function for which
    [tex]\frac d {dx} \exp(x) = \exp(x)[/tex]
    such that
    [tex]\exp(0) = 1[/tex].
    In this case, the answer is because its defined that way.

    2. The exponential function is defined by the power series
    [tex]\exp(x) = \sum_{n=0}^\infty \frac{x^n}{n!}[/tex]
    In this case, differentiating the power series that defines exp(x) yields the same power series.

    3. The exponential function is defined by the limit
    [tex]\exp(x) = \lim_{n\rightarrow\infty}\left(1+\frac x n\right)^n[/tex]
    In this case, differentiating the limit yields
    [tex]\exp(x) = \lim_{n\rightarrow\infty} \left(1+\frac x n\right)^{n-1}[/tex]
    Write the limit definition as
    [tex]\exp(x) = \lim_{n\rightarrow\infty}\left(1+\frac x n\right)^{n-1} + \lim_{n\rightarrow\infty}\frac x n\left(1+\frac x n\right)^{n-1}[/tex]
    The latter limit is clearly zero. The former limit is the same as the derivative of the defining limit. Thus differentiating the limit that defines exp(x) yields the same limit.
     
  17. Nov 18, 2005 #16
    [tex] y' = y[/tex]
    [tex]\frac{dy}{dx}=y[/tex]

    We multiply with [tex] dx [/tex] on each side and divide by [tex] y [/tex]:

    [tex] \frac{dy}{y} = dx [/tex]

    Integrals please!

    [tex] \int{\frac{dy}{y}} = \int{dx} [/tex]


    [tex]lny = x+C'[/tex]

    Out of knowledge of the natural logarithm:

    [tex]y = e^{x+C'}[/tex]
    [tex]y = e^xe^{C'}[/tex]
    [tex]C=e^{C'}[/tex]
    [tex]y = Ce^x[/tex]

    That's it! o:)
     
    Last edited: Nov 19, 2005
  18. Nov 19, 2005 #17
    Of course, without the knowledge that dex/dx=ex, you wouldn't know that the anti-derivative of 1/x is ln(x)!
     
  19. Nov 19, 2005 #18

    cepheid

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    It looks circular, but you could go the other way 'round the circle i.e. start by defining a function as a solution to the integral:

    [tex] f(x) = \int{\frac{dx}{x}} [/tex]

    Without evaluating the integral (because you don't know how), you can figure out some of the properties of the function and see that it has all the properties of a logarithm function (is there some way to make that more rigourous?) Call it the natural logarithm (it arose naturally in our investigation of that integral). Then ask, what is the base of this mysterious logarithm? What are the properities of the inverse exponential function? Once you find out that this base crops up everywhere in the natural sciences, it makes sense that it is natural exponential function.

    My first year calculus textbook did it both ways. Personally I like starting out with the exponential function (DH's post) better.
     
  20. Nov 20, 2005 #19

    The way we learned it when we studied engineering, was that [tex]ln(x)[/tex] is defined as the area between the vertical lines t=1, t=x, and between the function [tex]y=\frac{1}{t}[/tex] and the t-axis. (i.e., [tex]ln(x)=\int\limits_1^x{\frac{dt}{t}} [/tex])The constant [tex]e[/tex] is defined as the constant x that makes the area above exactly 1.

    Then, by studying this a little closer, we can see that the inverse of [tex]ln(x)[/tex] is [tex]e^x[/tex]. I don't remember quite exactly how we get to that, but that is at least why I wrote it in the post above. :rolleyes:
     
    Last edited: Nov 20, 2005
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