Why is the derivative of e^x also e^x?

Keith is wrong; the derivative of e^x is indeed e^x.

I can only offer a mathematical answer. It is because e^x is DEFINED to be the series

1+x+x^2/2! + x^3/3! + ...

Now if you just differentiate term by term you find that

d/dx ( 1+x+x^2/2! + x^3/3! + ...) = 1+x+x^2/2! + x^3/3! + ..

In other words, the derivative of e^x is e^x.

The definition of e^x as a series and theorems which say it is OK to differentiate this series term by term are from a part of mathematics known as power series.

Im not really sure how to say it but i think the easiest way to see is to use the taylor expansion of e^x=1+x+x^2/2!+x^3/3!+x^4/4!+....

clearly then if you differentiate e^x by differentiating its taylor series term by term you end with what you started with. Sorry I cant think of any better way of saying it!

Because the y-coordinate of y = e^x gives the slope of the tangent line?

Sounds a little like it's begging the question, doesn't it? Really, this is a phenomenon of mathematics, so you'll have to use mathematics (e.g. proof) to show why it is.

It's been a long time for me, but I don't think that's correct.

If y = e^x, then

dy/dx = xe^(x-1)