Earlier quoted context omitted.
That may be how it was arrived at historically, but it is not the best way to explain it. e arises when you ask the question: is there a function that is its own derivative? And it turns out the answer is yes. It is this infinite series: 1 + x + x^2/2 + x^3/6 + ... x^n/n! ... which you can easily verify is its own derivative simply by differentiating it term-by-term. When you evaluate this function at x=1, the result…
Another way to explain it, which I think is a "best" explanation in its own right, is that if you start differentiating exponential functions you can't help but discover e . The general differentiation of b^x, for b a constant and x the variable, is ln(x)*b^x. And of course to understand ln you need e . Or, to put it another way, you will blunder into this somewhere around month two of calculus 1, unavoidably. Of cou…
That's true, but it's begging the question because you can't define ln without already knowing about e.
You have to go back to first principles:
d(b^x) := lim(∂->0):(b^(x+∂) - b^x)/∂ = ((b^x)(b^∂) - b^x)/∂ = (b^x)(b^∂-1)/∂
So d(b^x) is itself multiplied by lim(∂->0):(b^∂-1)/∂. But now what? How do you evaluate that limit? How do you show that e is the magic value of b that makes that limit turn out to be 1? And in particular, how do you show that to someone whose only background knowledge is how to differentiate polynomials?
IMHO it's a lot easier to see that e is the value to which the polynomial series that is its own derivative converges at x=1.