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Definition of d(a^x)/dx

  1. Feb 6, 2006 #1

    G01

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    Hey. I'm having trouble understanding part of the definition of this derivative. Any help will be appreciated.

    [tex] f(x)=a^x [/tex]

    Using the definition of a derivative, the derivative of the above function is:

    [tex]f'(x) = \lim_{h \rightarrow 0}\frac{a^{x+h} + a^x}{h} = [/tex]

    [tex] \lim_{h \rightarrow 0} \frac{a^x(a^h - 1)}{h} [/tex]

    Since a^x does not depend on h it can be taken outside the limit:

    [tex] f'(x) = a^x \lim_{h \rightarrow 0} \frac{a^h-1}{h} [/tex]

    Now here is where I get confused. The text tells me that:

    [tex] \lim_{h \rightarrow 0} \frac{a^x(a^h - 1)}{h} = f'(0) [/tex] (1)

    If that is true then [tex] f'(x) = f'(0)a^x [/tex], but I have no idea why equation 1 is the way it is? How is that limit equal to f'(0)?:confused:
     
    Last edited: Feb 6, 2006
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  3. Feb 6, 2006 #2

    JasonRox

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    Note:

    [tex] \lim_{h \rightarrow 0} \frac{(a^h - 1)}{h} = \lim_{h \rightarrow 0} \frac{(a^{(0 + h)} - a^0)}{h} = f'(0)[/tex]
     
  4. Feb 6, 2006 #3

    AKG

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    You've made some errors. In the second line, you should have a minus sign, not a plus sign in the numerator. Equation (1) should read:

    [tex]\lim _{h \to 0}\frac{a^h - 1}{h} = f'(0)[/tex]

    You already have the equation:

    [tex]f'(x) = a^x\lim _{h \to 0}\frac{a^h - 1}{h}[/tex]

    Substitute 0 for x, and recognize that [itex]a^0 = 1[/itex], and you'll see why the equation for f'(0) holds.
     
  5. Feb 6, 2006 #4

    G01

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    Ahhh icic that was simpler than i thought. Thank you.
     
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