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I'm trying to develop a "natural sense" of calculus rather than just remembering formulas.

The proof of $de^x=e^x$ is rather interesting and I wonder if step (3) below means anything special? Is there an alternate, perhaps geometric, interpretation for it? Does it bear any special insight into why $e^x$ is such an important function in calculus?

(1) starting from the definition: $de^x = \lim_{\Delta x \to 0} \cfrac{e^{x+\Delta x} - e^x}{\Delta x}$

(2) after some algebra, I get: $d(e^x) = \lim_{\Delta x \to 0} \cfrac{e^x (e^{\Delta x} - 1)}{\Delta x}$

I noticed that $\lim_{\Delta x \to 0} \cfrac{e^{\Delta x} - 1}{\Delta x}$ is the the slope of $e^x$ when $x = 0$. so I cheat a little without going all the way to the definition of $e$ using limits.

(3) rewrite as $de^x = \lim_{\Delta x \to 0} e^x * \lim_{\Delta x \to 0} \cfrac{e^{\Delta x} - 1}{\Delta x}$

It seems to suggest that the derivative of $e^x$ is itself multiplied by the slope of itself when $x=0$ but saying so makes little sense. why is that? what am I missing?

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3 Answers

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You are completely right!

It turns out that $e^x$ is the only continous and differentiable function such that $f^\prime=f$ and $f(0)=1$. If we omit the last condition, then we get the class of all exponential functions - namely all functions $a^x$. You will see that you get the same calculations if you use $a^x$ instead of $e^x$. What is special about $e$ is precisely that the limit $$\lim_{h \to 0} \frac{{e^h}-1}{h}$$ is equal to one.

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"Does it bear any special insight into why $e^x$ is such an important function in calculus?" Yes. If you think of $\frac{d}{dx}$ as a linear operator that maps $f(x)$ to $f'(x)$, then it is reasonable to ask if such a linear operator has any invariant subspaces (i.e. $f'(x) = af(x)$) like matrices do. For matrices we call these invariant subspaces eigenvectors, for a linear operator like $\frac{d}{dx}$ we would call it an eigenfunction. Why this is so important is that in solving real-world problems? We often encounter (in physics and engineering) equations where an unknown function is a linear combination of its own derivatives. For example,

$$ f''(x) + af'(x) + bf(x) = 0 $$

It is clear that in order to solve it we need to find a function that does not fundamentally change when taking the derivative. It turns out that $e^{cx}$ is just that function. That is why $\frac{d}{dx}e^x = e^x$ is so important. It helps us solve practical problems.

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For any exponential function $a^x$, we get $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}a^x &=\lim_{h\to0}\frac{a^{x+h}-a^x}{h}\\ &=a^x\lim_{h\to0}\frac{a^h-1}{h}\\[6pt] &=c_a\,a^x \end{align} $$ Now, assuming the interchange of limits is okay, $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}e^x &=\frac{\mathrm{d}}{\mathrm{d}x}\lim_{n\to\infty}\left(1+\frac xn\right)^n\\ &=\lim_{n\to\infty}\frac nn\left(1+\frac xn\right)^{n-1}\\ &=e^x\lim_{n\to\infty}\left(1+\frac xn\right)^{-1}\\[6pt] &=e^x \end{align} $$ Therefore, $c_e=1$. In fact, the chain rule yields $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}a^x &=\frac{\mathrm{d}}{\mathrm{d}x}e^{x\log(a)}\\ &=\log(a)\,e^{x\log(a)}\\[6pt] &=\log(a)\,a^x \end{align} $$ Therefore, $c_a=\log(a)$.

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