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Is there a way to prove that the derivative of $e^x$ is $e^x$ without using chain rule? If so, what is it? Thanks.

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    do you know about Taylor series?2012-09-20
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    Taylor series assumes the existence of derivatives so this would be a circular argument.2012-09-20
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    @crf yea, but as DonAntonio said, it would be a circular argument2012-09-20
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    Why would you want to avoid using chain rule?2012-09-20
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    The chain rule is your friend...2012-09-20
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    @AlexBecker because the problem becomes trivial with chain rule, and I like thinking :)2012-09-20
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    @AlexBecker if there is no way to do it without chain rule, then thats fine. I was just wondering. Its been on my head for a while now.2012-09-20
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    What is your definition of $e^x$?2012-09-20
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    In my Analysis class, we defined $e^x$ as the solution of $f'(x) = f(x)$ with $f(0) = 1$. So um, that works.2012-09-20
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    How do you do this *with* the chain rule?2012-09-20
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    This question has been asked recently, but I can't find it...2012-09-20
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    @ChrisEagle let $y=e^x$ then $\ln(y)=x$ hence $\frac{1}{y}y'=1$ thus $y'=y$ aka $\frac{d}{dx}e^x=e^x$2012-09-20
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    @James: So defining $e^x$ as the inverse of the integral of $1/x$? How perverse.2012-09-20
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    $e^x=\cosh x + \sinh x, \cosh'=\sinh,\sinh'=\cosh$2012-09-20

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