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Derivative of a^x

  1. Jun 18, 2014 #1

    adjacent

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    I can't use the template here.
    Find the derivative of ##a^x##
    I know that it will be ##In(a).a^x##

    Now, I watched a lecture just now. How he derived this is as follows:
    $$\frac{\text{d}}{\text{d}x}a^x=a^x.\lim_{\Delta x \to 0}\frac{a^{\Delta x}-1}{\Delta x}$$

    (I omitted some steps). Then he made ##\lim_{\Delta x \to 0}\frac{a^{\Delta x}-1}{\Delta x}## a function defined as ##M(a)= \lim_{\Delta x \to 0}\frac{a^{\Delta x}-1}{\Delta x}##

    Do ##\frac{\text{d}}{\text{d}x}a^x=a^x.M(a)## .Now when we make x=0, the derivative of ##a^x## is equal to ##M(a)##.
    So ##M(a)## can be defined at the slope of the graph of ##a^x## when x=0.

    Then he introduced a new variable ##e## so that ##M(e)=1##
    That means ##\frac{\text{d}}{\text{d}x}e^x=e^x##
    But we don't yet know what ##e## is. We just defined it.
    Now, to find the derivative of ##a^x## what he did was:
    Convert it to base ##e##. ##a^x=e^{In(a)x}##

    I don't understand anything after this. Why did we define ##M(e)=1##? Why not ##M(e)=2##????
    And why ##e##? Why not ##z##?
    :confused:
     
  2. jcsd
  3. Jun 18, 2014 #2

    micromass

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    This are questions from a lecture, it is fine to post this in the main math forums. I'll move it to there.

    It's an L and not an I, so it's ln, not In. The ln is from logarithmus naturalis.

    These are just definitions. There is nothing to understand. He cold have chosen ##M(e) = 2## too, the theory would work perfectly with that choice. But historically, people have always taken ##M(e) = 1## and it's probably not a good idea to break with this history.
    One could also argue that we take ##M(e) = 1## because ##1## is simpler than ##2##.

    We could call it ##z## if we wanted to. It doesn't matter, it's just a name. But again, historically, we have give it the name ##e##. So we want to keep this tradition.
     
  4. Jun 18, 2014 #3

    adjacent

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    What if we took ##M(e)=0##? I doubt we can define it like that.

    Isn't ##e## the eulers number? This came from the study of compound interest? What does compound interest have to do with this ##e## we just defined?
     
  5. Jun 18, 2014 #4

    micromass

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    There is certainly some number ##a## such that ##M(a) = 0##. In fact, this number is ##1##. So ##M(1) = 0##.
    You can make any definition you like. The question is what we want to do with the definition. We have found that ##\frac{d}{dx} a^x = M(a)a^x##. This is true for all ##a##. It makes sense to see for which ##a## this equation is 'simplest', whatever that means. Well, you certainly can't argue that the equation takes a simple form if ##M(a) = 1##. So that looks promising. And indeed, the choice of ##a## such that ##M(a) = 1## turns out to be such a good choice that we have given it a special name, ##e##. Of course, this is posteriori knowledge. A priori, ##e## is just some number that makes the equation ##\frac{d}{dx} a^x = M(a)a^x## simple. It's only after figuring out more math that we really see its importance.

    Yes, it is in fact the same number. The relation to compound interest requires a proof that is not easy right now. I'm sure you will do the proof later.
    Although ##e## was first discovered with relation to this interest, its real importance is because of its uses in calculus.
     
  6. Jun 18, 2014 #5

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    Ok, I understand now. I have another question tho!
    How can we make ## \lim_{\Delta x \to 0}\frac{a^{\Delta x}-1}{\Delta x}## a function?
    Let's leave it and have another limit which is ##\lim_{\Delta x \to 0}2+\Delta x##
    Now this eventually becomes 2. How can we say that ##M(a)=\lim_{\Delta x \to 0}2+\Delta x## when ##\lim_{\Delta x \to 0}2+\Delta x## is always 2?
     
  7. Jun 18, 2014 #6

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    Hold on a second. Was one of those omitted steps the definition of ##a^x##?


    That choice leads to the *unique* function ##\exp(x)## such that ##\frac {d\exp(x)}{dx} = \exp(x)## and ##\exp(0)=1##.

    However, the approach taken begs the question, what does ##a^x## mean? A better way to look at it, at least to me, is to define the function ##\exp(x)## as satisfying ##\frac {d\exp(x)}{dx} = \exp(x)## and ##\exp(0)=1## and then proceed from there.
     
  8. Jun 18, 2014 #7

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    Every value of ##a>0## results in a specific value of that limit. The mapping from positive ##a## to values of the limit: That's a function.
     
  9. Jun 18, 2014 #8

    adjacent

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    No. He just used the definition of derivative.
     
  10. Jun 18, 2014 #9

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    How did he define ##a^x## then? He had to use some definition to be able to take that step from ##a^{x+\Delta x}-a^x = a^x\bigl(a^{\Delta x}-1\bigr)##.
     
  11. Jun 18, 2014 #10

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    Yeah, he said that ##a^x## is constant and he took it out of the limit.
    He said that ##\Delta x## is the thing that is moving, not x.
    I don't think I understood this very well
     
  12. Jun 18, 2014 #11

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    That step involves using ##a^{x+\Delta x} = a^x a^{\Delta x}##. How did he justify that step?
     
  13. Jun 18, 2014 #12

    micromass

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    You should probably get the book "First course in calculus" by Lang because he uses the same kind of argument to find the derivative of ##a^x##.
     
  14. Jun 18, 2014 #13

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    What he did is this:
    Using the definition of derivative,
    $$\frac{\text{d}}{\text{d}x}a^x=\lim_{\Delta x \to 0}\frac{a^{x+ \Delta x}-a^{\Delta x}}{\Delta x}$$
    Then using the laws of indices, this becomes:
    $$\lim_{\Delta x \to 0}\frac{a^xa^{\Delta x}-a^{\Delta x}}{\Delta x}$$
    Then,he factorised the numerator and got ##a^x(a^{\Delta x}-1)##

    Edit:
    I think you have told me that at least 10x :shy:
    I will buy it but my father is not at home these days. That's why I am watching the lectures first. :smile:
     
  15. Jun 20, 2014 #14
    He is probably working from the knowledge that the students have from precalculus. It is not rigorous, but the students have already seen the rules for ##a^x## when ##a>0## and ##x\in\mathbb{Q}##. I follow this general method in my AP class. We are answering the question, "assuming that ##a^x## is differentiable, what is the derivative?" The next step, after finding that the derivative is proportional to ##a^x## itself, is to wonder if there might be a value of ##a## such that ##\frac{d}{dx}a^x=a^x##.

    It seems reasonable in an introductory class to assume that ##a^x## follows the same rules for all ##a\in\mathbb{R}## and that the limit in question exists. I always state that these are assumptions that should be dealt with at some point, but not in my class.
     
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