Why is Napier's number, e=2.71828... so important?

I was recently reminded of the reason why Napier's number, \[e\approx 2.71828...\] is so important to calculus, and therefore most (if not all) of science and engineering, not to mention all the experiences/opportunities one may seek to understand as part of the human condition.

Simply put, the importance of e in these areas stems from the attempt at taking the derivative of an exponential function. That is, the derivative \[f'(x)=\frac{da^x}{dx}\]. Taking this derivative explicitly yields, \[\frac{da^x}{dx}=\lim_{\Delta x \rightarrow 0}\frac{a^{x+\frac{\Delta x}{2}}-a^{x-\frac{\Delta x}{2}}}{\Delta x}\]. The expression can be simplified by factoring out the common factor of \[a^x\], to get \[\frac{da^x}{dx}=a^{x}\lim_{\Delta x \rightarrow 0}\frac{a^{\frac{\Delta x}{2}}-a^{-\frac{\Delta x}{2}}}{\Delta x}\]. In addition, making use of the properties of limits, \[\frac{da^x}{dx}=a^{x}\left ( \lim_{\Delta x \rightarrow 0}a^{-\frac{\Delta x}{2}}\right )\left (\lim_{\Delta x \rightarrow 0}\frac{a^{\Delta x}-1}{\Delta x}\right )\] Where since, \[\left ( \lim_{\Delta x \rightarrow 0}a^{-\frac{\Delta x}{2}}\right )=1\], \[\frac{da^x}{dx}=a^{x}\left (\lim_{\Delta x \rightarrow 0}\frac{a^{\Delta x}-1}{\Delta x}\right )\], and\[\left (\lim_{\Delta x \rightarrow 0}\frac{a^{\Delta x}-1}{\Delta x}\right )=\frac{da^{x}}{d(x)}|_{x=0}=f'(0)\] one can see that taking this derivative might lead to a result that is counter intuitive. In fact, this seems kind of counter-intuitive and circular, since, according to this result, it would appear that one would need to know  the derivative of the exponential function to evaluate the derivative of the exponential function. \[\frac{d a^x}{dx}=f'(x)=f'(0)f(x)\]. One can get around this issue by looking for a solution where \[f'(0)=1\]. In fact, this condition is exactly what Napier's number yields. But don't just take anyone's word for it. This is provably so. That is, if for some number t you let, \[\frac{1}{t}=(a^{\Delta x}-1)\], then \[a^{\Delta x}=1+\frac{1}{t}\] and \[\Delta x = \log_a \left ( \frac{1}{t} + 1 \right ) \], and the question the becomes one where one should seek the base of the logarithm which yields unity. Since the previous limit can be assessed by approaching it from the positive and negative sides of the number 0. \[\lim_{\Delta x \rightarrow 0_\pm}\equiv \lim_{t\rightarrow\pm\infty}\]. In this way, one can relate the expression for the derivative of the exponential function to the limit of the number t via, \[\frac{da^x}{dx}=a^{x}\left (\lim_{\Delta x \rightarrow 0}\frac{a^{\Delta x}-1}{\Delta x}\right )=a^{x}\lim_{t\rightarrow \infty}\frac{1+\frac{1}{t}-1}{\log_a \left ( \frac{1}{t}+1\right )}\].

It would seem fortunate that since \[\lim_{t\rightarrow \infty}\frac{1+\frac{1}{t}-1}{\log_a \left ( \frac{1}{t}+1\right )}=\left [ \log_a \lim_{t\rightarrow \infty}\left (  \frac{1}{t}+1\right )^\frac{1}{t}\right ] ^ {-1}\] one can take advantage of the definition of Napier's number as the base of the logarithm to resolve this question, \[e=\lim_{t\rightarrow \infty}\left (  \frac{1}{t}+1\right )^\frac{1}{t}\] . 

Using these arguments, it is then abundantly clear that the derivative of any exponential function for any base is then explicitly dependent on Napier's number through the expression, \[\frac{da^x}{dx}=a^{x}\left [ \log_a \lim_{t\rightarrow \infty}\left (  \frac{1}{t}+1\right )^{\frac{1}{t}}\right ] ^{-1}\] which when written explicitly in terms of e is, \[\frac{da^x}{dx}=a^{x}\frac{1}{\log_a e}\]. 

To complete the story, for the case where a is defined to be 2.71828...=e, the derivative and the function are then provably equal: \[\frac{de^x}{dx}=e^{x}\frac{1}{\log_e e}= e^{x}\]