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Math Help - Derivative Proof for exponitials

  1. #1
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    Derivative Proof for exponitials

    Could someone please show me the proof that the derivative of F(x)=a(e)^x is F'(x)=a(e)^x?

    Thank you.
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  2. #2
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    This site explains what you need

    Derivative e^x

    also \frac{d}{dx}ae^x = a\frac{d}{dx}e^x =ae^x
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  3. #3
    Super Member Random Variable's Avatar
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     \frac{d}{dx} e^{x} = \lim _{\Delta x \to 0} \frac{e^{x+ \Delta x} - e^{x}}{\Delta x}

     = \lim _{\Delta x \to 0} \frac{e^{x}e^{\Delta x} - e^{x}}{\Delta x}

     = \lim _{\Delta x \to 0} \frac{e^{ x}(e^{\Delta x} - 1)}{\Delta x}

    by defintion,  e = \lim_{n \to \infty} \Big(1 + \frac{1}{n}\Big)^{n}

    let  n = \frac{1}{\Delta x}

    then  e = \lim_{\Delta x \to 0} (1 + \Delta x)^{1 / \Delta x}

    and  e^{\Delta x} = \lim_{\Delta x \to 0} (1 + \Delta x)


    finally  \frac{d}{dx} e^{x} = \lim _{\Delta x \to 0} \frac{e^{x} \Delta x}{\Delta x}   = e^{x}
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post

    then  e = \lim_{\Delta x \to 0} (1 + \Delta x)^{1 / \Delta x}

    and  e^{\Delta x} = \lim_{\Delta x \to 0} (1 + \Delta x)


    [/tex]
    This makes no sense.
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  5. #5
    Super Member Random Variable's Avatar
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    Quote Originally Posted by Drexel28 View Post
    This makes no sense.
    What exactly doesn't make sense?
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
    What exactly doesn't make sense?
    What you have said is analogous to saying that if m=\int_0^1 2^{\frac{1}{x}}dx then m^x=\int_0^1 2\text{ }dx. The \Delta x was a dummy variable, it was part of the limit. Outside of the limit it makes no sense.
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  7. #7
    Super Member Random Variable's Avatar
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    What if I did something as simple as reordering the steps?


    by defintion,  e = \lim_{n \to \infty} \Big(1 +  \frac{1}{n}\Big)^{n}

    let  n = \frac{1}{\Delta x}

    then  e = \lim_{\Delta x \to 0} (1 + \Delta x)^{1 / \Delta x}

    and  e^{\Delta x} = \lim_{\Delta x \to 0} (1 + \Delta x)


     \frac{d}{dx} e^{x} = \lim _{\Delta x \to 0} \frac{e^{x+ \Delta x} - e^{x}}{\Delta x}

     = \lim _{\Delta x \to 0} \frac{e^{x}e^{\Delta x} - e^{x}}{\Delta x}

     = \lim _{\Delta x \to 0} \frac{e^{ x}(e^{\Delta x} - 1)}{\Delta x} = \lim _{\Delta x \to 0} \frac{e^{x} \Delta x}{\Delta x}   = e^{x}
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
    What if I did something as simple as reordering the steps?


    by defintion,  e = \lim_{n \to \infty} \Big(1 +  \frac{1}{n}\Big)^{n}

    let  n = \frac{1}{\Delta x}

    then  e = \lim_{\Delta x \to 0} (1 + \Delta x)^{1 / \Delta x}

    and  e^{\Delta x} = \lim_{\Delta x \to 0} (1 + \Delta x)


     \frac{d}{dx} e^{x} = \lim _{\Delta x \to 0} \frac{e^{x+ \Delta x} - e^{x}}{\Delta x}

     = \lim _{\Delta x \to 0} \frac{e^{x}e^{\Delta x} - e^{x}}{\Delta x}

     = \lim _{\Delta x \to 0} \frac{e^{ x}(e^{\Delta x} - 1)}{\Delta x} = \lim _{\Delta x \to 0} \frac{e^{x} \Delta x}{\Delta x}   = e^{x}
    Still no cigar.
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  9. #9
    Super Member Random Variable's Avatar
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    OK. How about this:


     e = \lim_{n \to 0} (1 + n)^{1 /n}

    so for very small  n,  e \approx (1+ n) ^{1/ n}

    and  e^{n} \approx  (1 + n)


    which means  \lim_{n \to 0} \frac{e^{n}-1}{n} = \frac{n}{n} = 1


     \frac{d}{dx} e^{x} = \lim _{\Delta x \to 0} \frac{e^{x+ \Delta x} - e^{x}}{\Delta x}

     = \lim _{\Delta x \to 0} \frac{e^{x}e^{\Delta x} - e^{x}}{\Delta x}

     = e^{x}\lim _{\Delta x \to 0} \frac{e^{\Delta x} - 1}{\Delta x} = e^{x}(1) = e^{x}
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  10. #10
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
    OK. How about this:


     e = \lim_{n \to 0} (1 + n)^{1 /n}

    so for very small  n,  e \approx (1+ n) ^{1/ n}

    and  e^{n} \approx  (1 + n)


    which means  \lim_{n \to 0} \frac{e^{n}-1}{n} = \frac{n}{n} = 1


     \frac{d}{dx} e^{x} = \lim _{\Delta x \to 0} \frac{e^{x+ \Delta x} - e^{x}}{\Delta x}

     = \lim _{\Delta x \to 0} \frac{e^{x}e^{\Delta x} - e^{x}}{\Delta x}

     = e^{x}\lim _{\Delta x \to 0} \frac{e^{\Delta x} - 1}{\Delta x} = e^{x}(1) = e^{x}
    Better
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  11. #11
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    Random variables first proof makes quite a bit of sense to me. thanks for the proofs guys.
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