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0 pointsjerf1y ago0 comments

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 course, that doesn't show how it will show up in all sorts of other places; "the one and only function that is its own derivative" strikes me as more likely to be something we encounter everywhere.

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lisper1y ago

> The general differentiation of b^x, for b a constant and x the variable, is ln(x)*b^x.

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.

FabHK1y ago

b^x = (e^ln(b))^x = e^( ln(b) * x ),

and its derivative then is (by the chain rule)

ln(b) exp( ln(b) * x ) = ln(b) * b^x.

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