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If we define the number $e$ as $$e:=\lim_{n\to\infty}\left(1+{1\over n}\right)^n$$ then the only way I know to prove the derivatives of $e^x$ and it's inverse is to write $$\frac{\ln(x+h)-\ln x}{h}={1\over h}\ln\frac{x+h}{x}=\ln\left[\left(1+{h\over x}\right)^{1/h}\right]$$ and with some limit manipulations this can be shown to converge as $h\to 0$ to $$\ln(e^{1/x})={1\over x}$$ Now using the formula for the derivative of the inverse, $$\frac{d}{dx}e^x=\frac{1}{1/e^x}=e^x$$ Is there a way to get the derivative of $e^x$ directly from the definition of $e$ given above?

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  • $\begingroup$ The simplest approach to $\log$ and $\exp$ is through definition of $\log$ using integrals and defining $\exp$ as its inverse. If you insist on a limit based approach it is much better to define $\exp(x) = \lim\limits_{n \to \infty}\left(1 + \dfrac{x}{n}\right)^{n}$ and convince first that for rational $x$ we have $\exp(x) = \{\exp(1)\}^{x}$. Such an approach is provided at math.stackexchange.com/a/541330/72031 $\endgroup$ – Paramanand Singh May 4 '14 at 7:02
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Take 1: Are you familiar with the general result that if $f(x)$ is differentiable and has an inverse $g(x)$, that $g'(x)={1\over f'(g(x))}$?

Using that, we can prove ${d\over dx}e^x=e^x$ as follows. Let $f(x)=\ln x:=\int_1^x {1\over t}\,dt$ and let $g(x):=f^{-1}(x)=e^x$. Now $g'(x)={1\over f'(g(x))}$ and $$f'(x)=1/x\implies f'(g(x))={1\over e^x}\implies g'(x)={1\over {1/e^x}}=e^x.$$

(That didn't use your cited definition of $e$, but it accomplishes the task. Notice that doing it this way we don't presuppose any definition for $e^x$ other than it is the inverse function of $\ln x$.)


Take 2:

Consider $f(x)=a^x$, $a>0$. Then $$ f'(x)=\lim_{h\to 0}{f(x+h)-f(x)\over h}=\lim_{h\to 0}{a^{x+h}-a^x\over h}=\lim_{h\to 0}{a^x(a^h-1)\over h}=a^x \lim_{h\to 0}{a^h-1\over h}. $$ We want to find some value of $a$ such that $(a^x)'=a^x$, i.e. $\lim_{h\to 0}{a^h-1\over h}=1$. One way to define the number $e$ is that it is the unique number for which the last limit holds. Thus, $(e^x)'=e^x\cdot 1=e^x$.


Take 3: $$e:=\lim_{n\to\infty}\left(1+{1\over n}\right)^n\implies e^x=\left(\lim_{n\to\infty}\left(1+{1\over n}\right)^n\right)^x = \lim_{n\to\infty}\left(1+{1\over n}\right)^{nx}.\tag{$*$}$$

If $f(x)=e^x$ (as defined above), then $$ f'(x)=\lim_{h\to 0}{e^{x+h}-e^x\over h}=e^x\lim_{h\to 0}{e^h-1\over h}, $$ and thus, with this approach, your question boils down to showing that $$ \lim_{h\to 0}{e^h-1\over h}=1 $$ where---and this is the key---we are working with $e^h$ as defined in $(*)$.

Typically, you show this by appealing to the infinite series definition of $e^x$ and the Binomial Theorem. However, if we want to work from $(*)$, we can do this:

Let $n=1/h$ in $(*)$ so $\displaystyle e=\lim_{h \to 0}(1+h)^{1/h}\implies e^h\approx 1+h$ when $h\approx 0$. Thus, $$ \lim_{h\to 0}{e^h-1\over h}=\lim_{h\to 0}{(1+h)-1\over h}=1. $$


Take 4: A classical approach is to define the function $e^x$ as the unique solution to the initial value problem ${dy\over dx}=y$, $y(0)=1$. Of course the existence and uniqueness here must be proven, not simply asserted, but this requires more work than I want to type and is available in any standard reference on ODE theory.

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  • $\begingroup$ I want $e^x$ defined in terms of the number $e$. Otherwise your symbol $e^x$ has nothing to do, a priori, with that number since it was defined as the inverse of an integral. But that's a very nice approach, thanks for the post! I like having many different ways to prove things. $\endgroup$ – user142299 May 4 '14 at 1:48
  • $\begingroup$ Do you specifically want that definition of the number $e$ (after which the function $e^x$ makes sense), or will another (equivalent) numerical definition suffice? $\endgroup$ – JohnD May 4 '14 at 1:57
  • $\begingroup$ That always seems the easiest to motivate to me, since it comes out of simple limit considerations (such as compound interest problems). If you're referring to the series definition, then no I would prefer not. But like I said I enjoy seeing different approaches, so if in doubt post. $\endgroup$ – user142299 May 4 '14 at 1:59
  • $\begingroup$ How do you establish that $\lim \frac{e^h-1}{h}=1$? $\endgroup$ – user142299 May 4 '14 at 4:21
  • $\begingroup$ Typically by using $e=\sum_{n=0}^\infty {1\over n!}$ and the Binomial Theorem. But I've added a different argument to the end of Take 3. $\endgroup$ – JohnD May 4 '14 at 4:44

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