Deriving e^x

Why is the derivative of e^x also e^x? I don't understand...

Can someone post a proof?

e^x = y
Rewrite using natural log
ln y = x
Implicitly differentiate both sides
dx(ln y) = dx(x)
A bit of algebra
(1 / y)(dy/dx) = 1
dy/dx = y
dy/dx = e^x

Hope that helps.

>>2 That’s a proof defining e to be the inverse function of natural log.

Alternatively define e to be a number such that lim h->0 (e^h–1)/h = 1.  Let f(x) = e^x
f’(x) = lim h->0 (f(x+h)-f(x))/h
= lim h->0 ((e^x)(e^h)-e^x)/h
= (e^x)(lim h->0 (e^h-1)/h)
= e^x

Basically, e is defined such that e^x has the derivative e^x.

>>4
 got it right. it's just a tautology.

Or e = lim:n->inf(1 + 1/n)^n

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