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Derivative of a^x using limit definition

  1. Oct 6, 2011 #1
    1. The problem statement, all variables and given/known data

    Sketch and label on the same pair of axes the graphs of y=f(x) and y=f'(x) for ... c) f(x)=2x

    2. Relevant equations

    3. The attempt at a solution

    f(x) = 2x
    f'(x) = lim as h→0 (2x+h-2x)/h
    = lim as h→0 (2x2h-2x)/h
    = lim as h→0 2x(2h-1)/h
    = lim as h→0 2x ∙ lim as h→0 (2h-1)/h
    = 2x ∙ lim as h→0 (2h-1)/h

    I know I have to show that the limit after 2x is equal to ln(2), but how do I do that using this definition?
    Thanks.
     
  2. jcsd
  3. Oct 6, 2011 #2

    Ray Vickson

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    I have a question that looks frivolous but is actually serious. What do you mean by 2^x? You know how to define 2^n for positive or negative integers n and how to define 2^r for rational r of the form k/2^n, but what do you mean by 2^x? Most definitions would use logarithms to define it, or else use some limiting process, etc. Related questions: how do you know that 2^x is differentiable? All these issues are easiest if you use logarithms. In that case the answer falls out almost without effort.

    RGV
     
  4. Oct 6, 2011 #3
    I apologize if my answer doesn't satisfy your question, but by 2x I mean the graph that represents 2 to the power of any real number.

    As for your relevant question, I know 2x is differentiable because it's continuous while also being a smooth curve with no vertical tangents. I also have previous experience in deriving any base to the power of x, but I'm expected to derive 2x using the limit definition.
     
  5. Oct 6, 2011 #4

    Ray Vickson

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    You have completely misunderstood my question. I know what 2^x is supposed to be. My question is more fundamental than that: how do you define it? We know how to define 2^1, 2^2, 2^3, ... and 2^(-1), 2^(-2),..., as well as 2^(1/2), 2^(1/4), 2^(3/4), etc. But how do we define 2^x for general, real values of x? Of course we all know that f(x) = 2^x is continuous, differentiable, etc., but why do we know that? Well, in my case it was because f(x) was defined in an unambiguous way in terms of logarithms and exponentials, from which those properties follow easily. If you define it in some other way, you are left with the thorny problem of actually *proving* what you say; alternatively, you could handwave and hope for the best.

    RGV
     
  6. Oct 6, 2011 #5
    2h = ehln2.

    What happens if you write the series expansion for it?
     
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