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Math Help - proving the exponential function is differentiable

  1. #1
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    proving the exponential function is differentiable

    Hi my question is :


    Show that e^{x}is differentiable on \mathbb{R}


    I began by the definition that A function f:A\rightarrow\mathbb{R}is differentiable at c if and only if

    \underset{x\rightarrow c}{lim}\frac{f(x)-f(c)}{x-c}

    I started by  \underset{x\rightarrow c}{lim}\frac{e^{x}-e^{c}}{x-c}however i am now unsure on how to cancel the x-c bit.

    I am not sure on how to show this.

    Thanks
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  2. #2
    Senior Member roninpro's Avatar
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    It might be better to consider the alternate definition:

    \displaystyle f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}

    Then you would have to compute

    \displaystyle \lim_{h\to 0} \frac{e^{x+h}-e^x}{h}=e^x \lim_{h\to 0} \frac{e^h-1}{h}

    (Handling the remaining limit depends on how you define e.)
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by cooltowns View Post
    Hi my question is :


    Show that e^{x}is differentiable on \mathbb{R}


    I began by the definition that A function f:A\rightarrow\mathbb{R}is differentiable at c if and only if

    \underset{x\rightarrow c}{lim}\frac{f(x)-f(c)}{x-c}

    I started by  \underset{x\rightarrow c}{lim}\frac{e^{x}-e^{c}}{x-c}however i am now unsure on how to cancel the x-c bit.

    I am not sure on how to show this.

    Thanks
    In...

    http://www.mathhelpforum.com/math-he...tml#post582679

    ... it has been extablished that f(x) is differentiable in x=a if there exist a real constant k such that...

    \displaystyle f(x) = f(a) + k\ (x-a) + o(x-a) (1)

    Now if You remember the 'fundamental limit'...

    \displaystyle \lim_{x \rightarrow a} \frac{e^{x-a}-1}{x-a} = 1 (2)

    ... You from (2) derive that is...

    \displaystyle e^{x-a} -1 = (x-a) + o(x-a) \implies \frac{e^{x}}{e^{a}} = 1+(x-a) + o(x-a) \implies e^{x} = e^{a} + e^{a}\ (x-a) + o(x-a) (3)

    ... so that (1) is verified...

    Kind regards

    \chi \sigma
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