Proving derivative of $e^x$ and $\ln x$

Question

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?

Answer 1

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 $\ln x$.)

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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, $(e^x)'=e^x\cdot 1=e^x$.

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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, $$ \lim_{h\to 0}{e^h-1\over h}=\lim_{h\to 0}{(1+h)-1\over h}=1. $$

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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.