Why is $\frac{d}{dx}\exp(x) = \exp(x)$?

Question

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?

Answer 1

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.)

Answer 2

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 $0$.

Answer 3

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 $x = 1$.

Answer 4

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).