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Question about the derivative of e^x

  1. Oct 6, 2008 #1
    I've been drawn to this expression for a long time.

    Anyway, my question is - is the derivative of e^x exact? That is with absolute precision, the derivative of e^x is EXACTLY e^x?

    Sorry if this seems like a silly question, I would really like to know if my math instructor is right on this one

    I remember the argument:

    lim h-->0 e^x+h - e^x/ h

    What about the constant e? Is it approximate or absolutely precise?

    Say you have a base, b.

    Then lim q->0 ([b^1+q -b]/q) = b, which through some adjustments gives e = lim n-> infinity (1 +1/n)^n
    Last edited: Oct 7, 2008
  2. jcsd
  3. Oct 7, 2008 #2


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    Yes. In general, when you derive [itex]g^x[/itex], with g any number, the derivative is
    [tex]\operatorname{ln}(g) g^x = k \log(g) g^x[/tex], where "ln" is some weird function which turns out to be the logarithm (in base 10), up to a multiplicative constant k.
    Now when you plug in some numbers for g, you will find that this function "ln" comes close to unity if you take g to be somewhere around 2.7. Now it would be very cool to know this number g for which g^x is its own derivative exactly, but it turns out not to be anything nice (for example, it is not a fraction, or a square root of something). So we define e to be this number, for which
    [tex]\operatorname{ln}(e) = 1[/tex]
    and then of course (by the properties of the logarithm) the constant k is
    [tex]k = 1/\operatorname{{}^{10}log}(e)[/tex].

    So, by (one the many equivalent) definition of e, the derivative of e^x is e^x. As long as you write e and not some numerical approximation like 2.71828... this is an exact identity. Indeed, by definition of the derivative one can show that this requirement of e^x being its own derivative is equivalent to using
    [tex]e \stackrel{\text{def}}{\equiv} \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^n[/tex].
  4. Oct 7, 2008 #3


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    Yes, the number e, like any number, is exact. The derivative of the function ex is exactly ex. Pretty nice function, eh?
  5. Oct 7, 2008 #4
    The coolest function ever defined in calculus is ex. You differentiate or integrate this function you get back the function. Who actually the first to defined such uninteresting function?
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