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# $$e$$

The derivative of $$e^x$$ is itself. That's how $$e$$ is (maybe I should say 'can be') defined.

Lets do some math with that. By the above definition, we have $$e^x={e^{x+dx}-e^x\over dx}$$ where, of course, $$dx$$ goes to 0. We can now simplify this to $$dx=e^{dx}-1$$. Hey! We can solve for $$e$$! Lets do that: $$e=(1+dx)^{1\over dx}$$. Since $$dx$$ is going to 0, we can replace it by, for example, $$1\over n$$, where $$n$$ goes to infinity. We get the well-known expression$e=\lim_{n\rightarrow\infty}\left(1+{1\over n}\right)^{n}$

Just because it's fun to do, I'll also give you another way to approach $$e$$. Taylor series are always fun, so consider the following: $e^x=\sum_{k=0}^{\infty}{a_kx^k}$Since the $$n$$th derivative on the left at $$x=0$$ is always 1, the right side is also 1; $$1=n!a_n$$. Thus, $$a_n={1\over n!}$$, so we get $e^x=\sum_{k=0}^{\infty}{x^k\over k!}$Note that this is not a full proof, but it gives you an idea of what happens. You can also make a Taylor series for $$\sin(x)$$ and $$\cos(x)$$, and with those, you can see that $$e^{ix}=\cos(x)+i\sin(x)$$, where $$i^2=-1$$ is the imaginary number. This connects $$\pi$$ to $$e$$. Awesome, right?

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