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Calculating the definition of a derivative: 2^x

  1. Dec 5, 2014 #1
    1. The problem statement, all variables and given/known data

    I'm supposed to find the derivative of 2^x using the definition of a derivative. I am really confused as to how I can factor out the h.

    2. Relevant equations

    3. The attempt at a solution
    limit as h->0 in all of these, I don't want to write it out because it's going to look even worse (didn't want to learn latex right this moment).
    This is as far as I got and I'm not even sure if it was useful to factor out 2^x
    Hopefully there is a rule for exponents that I'm not aware of that can be used for this.
  2. jcsd
  3. Dec 5, 2014 #2


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    Use the fact that ##2^h = e^{h \log 2}## (where ##\log## is the natural log).

    I guess you need to know some other stuff as well. Without knowing what you're allowed to use, it's hard to give you advice. Do you know the derivative of ##e^x## for example? Or the following limit?
    $$\lim_{h\to 0}\frac{e^h - 1}h = 1$$
  4. Dec 6, 2014 #3
    Honestly I'm not entirely sure what I'm allowed to use, we went over the proof for the derivative of e but we were told that we wouldn't need to know it, this question is from a previous year so maybe that year the curriculum was different? I know how to derive e^x using the chain rule but I'm not sure how I could incorporate that into my limit.
  5. Dec 6, 2014 #4


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    Well, that's one thing you'll need to figure out.

    You can solve this problem pretty easily if you can use the fact that the derivative of ##e^x## is ##e^x##. You don't need to know how to prove it.

    I think you mean you know how to differentiate ##e^x##. Derive doesn't mean "find the derivative of."
  6. Dec 6, 2014 #5


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    "Using the definition of a derivative" means using the difference quotient:
    [tex]\frac{df}{dx}= \lim_{h\to 0} \frac{f(x+ h)- f(x)}{h}[/tex]

    [tex]\frac{2^{x+h}- 2^x}{h}= \frac{2^x(2^h)- 2^x}{h}= 2^x \frac{2^h- 1}{h}[/tex]

    So [itex]\frac{d 2^x}{dx}= 2^x\left(\lim_{h\to 0} \frac{2^h- 1}{h}\right)[/itex]
    That is obviously [itex]2^x[/itex] times a constant. How to find that constant, and if you can, depends on what other knowledge you have.
    ([itex]2^x= e^{ln(2^x)}= e^{x ln(2)}[/itex] so if you know how to differentiate [itex]e^x[/itex] you can show that constant is ln(2).)
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