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The Number e and Logarithms

  1. Mar 2, 2005 #1


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    The Number "e" and Logarithms

    Hi, I'm having a lot of difficulty understanding the number "e" and logarithms, especially in terms of differentiating them. Is it just a matter of memorization or are there tricks to finding the derivatives? Help!
  2. jcsd
  3. Mar 2, 2005 #2


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    Yes,u better memorize them.It's much better than deducing it with every occasion (:yuck:)...

    Thery're the simplest diff.rules possible.

  4. Mar 2, 2005 #3


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    There are different ways of defining and introducing e, but when you're new to logarithms, I think the easiest way is to define e as that number for which:

    More detail:

    [tex]\frac{d}{dx}a^x=\lim_{h \to 0} \frac{a^{x+h}-a^x}{h}=a^x\lim_{h \to 0}\frac{a^h-1}{h}[/tex]

    You can check the limit exists (for a>0 ofcourse). Also, if [itex]f(x)=a^x[/itex], then you can see that [itex]f'(x)=a^xf'(0)[/itex]
    If you try some values for a and some small values for h to get some sight as to the value of the limit :

    If a=2, then the limit is approx. 0.6934
    If a=3, then the limit is approx. 1.0986

    The larger a, the larger the limit. There exists some number between 2 and 3 for which is value is 1. You can define that number to be e.
    So e is that number for which:
    [tex]\lim_{h \to 0}\frac{e^h-1}{h}=1[/tex]

    Therefore you have this nice rule when differentiating e^x:
    and this is actually one of the main reasons it is used so often.

    Likewise, differentiating [itex]\log_a x[/itex] gives [itex]\frac{1}{x \ln a}[/itex], so if you use base e (the natural logarithm) the derivative is simplified.

    The above hopefully gives some insight into e. It's important to understand where it comes from and why it is used so much. After that, memorizing the derivatives and such is trivial.
    Last edited: Mar 2, 2005
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