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Derivative of the exponential function with base e

  1. Jul 8, 2012 #1
    For an exponential function of the form y=a^x
    First derivative , d/dx [a^x ] = a^x∙ lim┬(δx→0)⁡{(a^δx-a^0)/δx}
    = a^x∙ m_((0,1))
    now if m_((0,1)) which is the gradient of the y-axis intersection point of the exponential function equal to 1 exactly, then the derivative will be unchanged under differentiation and the value of the base a is then considered as e (Napier constant/natural number) .
    IS there any derivation to get to the napier constant??
  2. jcsd
  3. Jul 8, 2012 #2


    Staff: Mentor

  4. Jul 8, 2012 #3


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    If I understand your question, one way is this: for [itex]f(x)= a^x[/itex] we have
    [tex]\frac{f(x+h)- f(x)}{h}= \frac{a^{x+h}- a^x}{h}= \frac{a^xa^h- a^x}{h}= a
    ^x\frac{a^h- 1}{h}[/tex]

    Now, if we take the limit as h goes to 0, we have
    [tex]\frac{da^x}{dx}= a^x\lim_{h\to 0}\frac{a^h- 1}{h}[/tex]

    So we will have the very nice property that [itex]de^x/dx= e^x[/itex] if and only if
    [tex]\lim_{h\to 0}\frac{e^h- 1}{h}= 1[/tex]

    Now, turn that around. I will just give a rough argument here- for h very close to 0, that limit says that [itex](e^h- 1)/h= 1[/itex], approximately. Then [itex]e^h- 1= h[/itex], approximately. From that, [itex]e^h= 1+ h[/itex], approximately, so that [itex]e= (1+ h)^{1/h}[/itex]. If we let n= 1/h, n goes to infinity as h goes to 0 and so the approximation [itex]e= (1+ 1/n)^n[/itex] becomes exact as n goes to infinity:
    [tex]e= \lim_{n\to\infty}\left(1+ \frac{1}{n}\right)^n[/tex]
    the "standard" limit definition of n.

    Personally, I prefer to start by defining [itex]ln(x)= \int_1^x dt/t[/itex], then defining exp(x) to be the inverse function to ln(x), finally showing that [itex]exp(x)= exp(1)^x= e^x[/itex] where e is now defined to be the value of x such that ln(x)= 1.
  5. Jul 12, 2012 #4
    Thanks a lot! you really inspiring me!
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