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.
Proof of derivative of $e^x$ is $e^x$ without using chain rule
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$\begingroup$
calculus
derivatives
exponential-function
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0do you know about Taylor series? – 2012-09-20
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0Taylor series assumes the existence of derivatives so this would be a circular argument. – 2012-09-20
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0@crf yea, but as DonAntonio said, it would be a circular argument – 2012-09-20
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0Why would you want to avoid using chain rule? – 2012-09-20
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0The chain rule is your friend... – 2012-09-20
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0@AlexBecker because the problem becomes trivial with chain rule, and I like thinking :) – 2012-09-20
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0@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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25What is your definition of $e^x$? – 2012-09-20
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3In 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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7How do you do this *with* the chain rule? – 2012-09-20
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0This question has been asked recently, but I can't find it... – 2012-09-20
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4@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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1@James: So defining $e^x$ as the inverse of the integral of $1/x$? How perverse. – 2012-09-20
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0$e^x=\cosh x + \sinh x, \cosh'=\sinh,\sinh'=\cosh$ – 2012-09-20