Why the slope of $e^x$ at any point is it's value at that point? I know that we can differentiate $e^x$ by using limits. But why is it that it is the derivative of it's own? Or is it the case that there must exist a function which should be it's own derivative and we have defined it to be $e^x$?
 A: There is an excellent book, "e:" The Story of a Number, that goes through the many ways $e$ and $e^x$ and $\ln(x)$ can be defined, and the historical motivations for the development of that math.
A bit surprisingly, the historical reasoning was:


*

*The integral of the function $f(x) = \frac1x$ exists and $\int_1^t \frac1x\,dx$ is a single-valued function of the upper limit  $t$.  Let's name that function $\log(t)$.

*You can prove some interesting things about $\log(t)$ including the familar property that $\log(uv) = \log(u)+\log(v)$ just from the definition as an integral (!)

*You can easily prove that $\log(x)$ is monotonic increasing and continuous, so there is an inverse function, which for now we can call $\exp(x)$.

*Using the properties of $\log(x)$ you can prove that $\exp(x)$ has the usual properties of some power $a^x$, for example, $\exp(x)\exp(y) = \exp(x+y)$.

*If we identify the base of that power as some specific number, call it $e$, we can then prove some amazing properties of $e$, including that
$$
\frac{d(e^x)}{dx} = e^x\\
e = \lim{x\to\infty} \left( 1+\frac1x\right)^x \\
e^x = \sum_{n=0}^\infty \frac{x^n}{n!}
$$
The proof that $\frac{d(e^x)}{dx} = e^x$ is particularly easy, because $\frac{d(\log(x))}{dx} = \frac1x$ by the definition of $\log(x)$, and $e^x$ is the inverse function of $\log(x)$.
The book, by the way, also tells you why "log" is related to the word "log" meaning a piece of a tree.  I won't spoil it by revealing the relation; you will love to read the book.
