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July 10th, 2019, 07:28 AM  #1 
Senior Member Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8 
People prove the derivative of $\displaystyle e^x$, $\displaystyle \ln(x)$ using either the formula of $\displaystyle e = \lim_{x \to 0} (1+x) ^ \frac{1}{x}$ or if they know one of the derivatives $\displaystyle e^x$, $\displaystyle \ln(x)$ they use it to prove the other. My textbook first gave me without proof that $\displaystyle \frac{d}{dx}e^x = e^x$ Then it found the limit of $\displaystyle \ln(x) $ using the chain rule: $\displaystyle f(x) = \ln(x) \Leftrightarrow e^{f(x)} = e^{\ln(x)} \Leftrightarrow e^{f(x)} = x \Leftrightarrow \frac{d}{dx}e^{f(x)} = \frac{d}{dx}x \Leftrightarrow e^{f(x)} \frac{df(x)}{dx} = 1 \Leftrightarrow \frac{df(x)}{dx} = \frac{1}{e^{f(x)} } \Leftrightarrow \frac{d}{dx}\ln x = \frac{1}{e^{\ln x} } \Leftrightarrow \frac{d}{dx}\ln x = \frac{1}{x} $ And then it uses the limit of ln(x) at x=1 as calculated above with the definition of the derivative in order to show that $\displaystyle e = \lim_{x \to 0} (1+x) ^ \frac{1}{x}$ And then others (on the internet) are using $\displaystyle e = \lim_{x \to 0} (1+x) ^ \frac{1}{x}$ to show that the derivative of $\displaystyle e^x$ is $\displaystyle e^x$. Well the circular problem is solved if you can some how find the derivative of $\displaystyle e^x$ without using any of the above. Then all the other derivatives and e as a limit can be proved by using $\displaystyle e^x$. Can you show me the right order? Well the book might have a good reason that it hasn't shown to me yet; maybe it's really difficult to calculate the limit of $\displaystyle e^x$. Thank you. Last edited by skipjack; July 10th, 2019 at 11:06 AM. 
July 10th, 2019, 08:32 AM  #2 
Senior Member Joined: Oct 2009 Posts: 841 Thanks: 323 
If you want to avoid circular reasonings, you'd need to start by adressing what is the DEFINITION of e, e^x and log(x).

July 10th, 2019, 08:37 AM  #3 
Senior Member Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8  
July 10th, 2019, 08:56 AM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 2,500 Thanks: 1372  
July 10th, 2019, 09:06 AM  #5 
Senior Member Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8  
July 10th, 2019, 09:30 AM  #6 
Senior Member Joined: Sep 2015 From: USA Posts: 2,500 Thanks: 1372  A bit of reading on the web will provide you with all the answers to this question that exist.

July 10th, 2019, 11:38 AM  #7 
Global Moderator Joined: Dec 2006 Posts: 20,823 Thanks: 2159  One needs to start somewhere. For $x > 0$, it defines a function of $x$ that can be called "$\ln(x)$" and has the properties you'd like for a logarithm. One can now define $e$ as the value of $x$ such that $\ln(x) = 1$. 

Tags 
circular, dependency, limit, lnx, proof 
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