Why is $\frac{d}{dx}\exp(x) = \exp(x)$? What is the explanation for
$$(e^x)'=e^x$$
I searched the SE, 'cause this can't be the first time this has been asked. But the question seems hard to formulate and search for here and on Google. Any help?
 A: Note that for $f(x) = e^x$ you have by definition
$$
f'(x) = \lim_{h\to 0} \frac{e^{x+h} - e^x}{h} = \lim_{h\to 0} \frac{e^{x}(e^h - 1)}{h} = e^x\lim_{h\to 0} \frac{e^h - 1}{h}.
$$
Now it all comes down to how you show that
$$
\lim_{h\to 0} \frac{e^h - 1}{h} = 1.
$$
And this will depend on how you have defined $e^x$ in the first place.
Lets say we have defined
$$
e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}.
$$
Then
$$
\frac{e^h - 1}{h} = \sum_{n=1}^{\infty} \frac{h^{n-1}}{n!} = 1 + \sum_{n = 2}^{\infty} \frac{h^{n-1}}{n!} \to 1 \text{ as } h\to 0.
$$
(You might have to think about this limit a bit.)
A: If you consider $e^x = \exp(x)$ to be the function defined by the power series $\sum_{n=0}^{\infty} \frac{x^n}{n!}$ (where I assume you're talking about $x \in \mathbb{R}$ or $x \in \mathbb{C}$), then you can show that the radius of convergence of this power series is $\infty$. This let's you commute the sum and the derivative for $x$ in any domain in $\mathbb{R}$ or $\mathbb{C}$ to find
$$
\frac{d}{dx}\exp(x) = \frac{d}{dx} \sum_{n=0}^{\infty} \frac{x^n}{n!} = \sum_{n=0}^{\infty} \frac{d}{dx} \frac{x^n}{n!} = \sum_{n=1}^{\infty} \frac{n x^{n-1}}{n!} = \sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1)!} = \sum_{n=0}^{\infty} \frac{x^n}{n!} = \exp(x).
$$
Note that along the way the sum starting from $n=0$ switched to $n=1$ because the $n=0$ term is constant and therefore its derivative is $0$.
A: The proof depends on the way you define $e$, but I like the following theorem as the motivation for considering $e$ at all.
Theorem. Let $a > 0$ and consider the function $f_a(x) = a^x$. Then $f_a'(x) = c_a a^x$ for some $c_a \in {\mathbb R}$.
Proof. $(f_a(x + h) - f_a(x))/h = (a^{x+h} - a^x) / h = a^x (a^h - 1) / h$. Take $c_a$ to be $\lim_{h \to 0} (a^h - 1)/h$. (For brevity, I'm skirting over the existence of this limit). Then $f_a'(x) = c_a a^x$. $\Box$
Of course, given $a$, the $c_a$ from the previous theorem is unique; and the proof also tells what $c_a$ is, namely $c_a = \lim_{h \to 0} (a^h - 1)/h$. Now it's quite natural to notice that a number $a$ for which $c_a$ happens to be 1 is interesting, because that happens to have the property that $f_a' = f_a$. That number is defined to be $e$. (Again, I' skipping some arguments, namely that there is an $a$ for which $c_a = 1$ and that there is only one). To see that this is the $e$ we all know and love as $\sum_{n=0}^\infty 1/n!$, you can now consider $f_e(x) = \sum_{n=0}^\infty f_e^{(n)}(0)x^n/n! = \sum_{n=0}^\infty f_e(0)x^n/n! = \sum_{n=0}^\infty x^n/n!$ at $x = 1$.
A: This may be convoluted way of going about the problem. I think of it as the following:
Given that $$\log(e^{x})=x$$
If we were to differentiate both sides w.r.t. x, we get,
$$\frac{1}{e^{x}}\frac{d}{dx}e^{x}=1$$
Now, let's call $\frac{d}{dx}e^{x}=(e^{x})'=y$. Then, $\frac{y}{e^{x}}=1 \Rightarrow y= e^{x}$
So, $(e^{x})'$ has to be equal to $e^{x}$.
Of course, to get to that you'll have to know the derivative of log(x).
