Logarithms and Exponentials
Some textbooks will define $e$ by its derivative, others define it by its Taylor expansion. For this series, I am going to use the limit definition.
\[e^x = \lim_{n \rightarrow \infty} \left ( 1 + \frac{x}{n} \right )^n\]Logarithms
I will assume the reader is comfortable with the properties of logarithms. Also, this proof uses the fact that logarithms are continuous. I prove this in this post of my limits and continuity series.
Let $b \in \mathbb{R}$ and $b > 1$.
\[\begin{align} \frac{d}{dx} \log_b (x) &= \lim_{h \rightarrow 0} \ \frac{\log_b(x+h) - \log(x)}{h} \\[10pt] &= \lim_{h \rightarrow 0} \ \frac{1}{h} \log_b((x+h)/x) \\[10pt] &= \lim_{h \rightarrow 0} \ \log_b \left ( \left ( 1 + \frac{h}{x} \right )^{1/h} \right ) \\[10pt] &\text{let } n = x/h \text{ therefore } n \rightarrow \infty \text{ as } h \rightarrow 0 \text{ since } x \text{ is fixed and finite} \\[10pt] &= \lim_{n \rightarrow \infty} \ \log_b \left ( \left ( 1 + \frac{1}{n} \right )^{n/x} \right ) \\[10pt] &= \log_b \left ( \lim_{n \rightarrow \infty} \ \left ( 1 + \frac{1}{n} \right )^{n} \right )^{1/x} \\[10pt] &= \log_b \left ( e \right )^{1/x} \\[10pt] &= \frac{1}{x} \log_b \left ( e \right ) \\[10pt] &= \frac{1}{x \ln b} \end{align}\]In particular, if we let $b = e$, then
\[\frac{d}{dx} \ln (x) = \frac{1}{x}\]Exponentials
Now, we use the fact that $\log_b (b^x) = x$ and the inverse function derivative property.
\[\begin{align} \frac{d}{dx} b^x \quad=\quad \frac{1}{\frac{d}{dz} \log_b (z) \bigg\rvert_{z=b^x} } \quad=\quad \frac{1}{\frac{1}{z \ln b} \bigg\rvert_{z=b^x} } \quad=\quad b^x \ln b \end{align}\]In particular, if we let $b = e$, then
\[\frac{d}{dx} e^x = e^x\]This is what makes $e$ such a special constant. It represents the fixed-point of the derivative operation. In fact, it is the unique fixed point. Thus, if another function is found to also be a fixed-point, then it must be equal to $e^x$. Maybe you should compute the derivative of $f(x) = \cos (x/i) + i \sin (x/i)$ where $i$ is an imaginary number with the property that $i^2 = -1$.