feanur is right


you're saying (f(g))' = f(g)*g', i suggest you read a bit more on the chain rule

It's defined that way, hence why we have the constant "e".

It is THE sole function who, when differentiated, produces its own derivative. (Advanced calculus/analysis will actually take this as an axiom and prove it is true and unique).

I know there are smart asses out there, so I'll point out that f(x) = 0 does NOT count.

There's more ways to define e than there are.........(insert analogy here).

Calculus is your friend here if you really want to know...derivatives happen to "match up" when the base happens to be 2.17.... or "e"

This post was edited by Casey on Jun 20 2015 08:55pm

It's been 5 years since I learned this but I was taught it the following way.

dy/dx = y
dy/y = dx
Int (1/y) dy = Int (1) dx
Ln(y) = x
y = e^x

I was sloppy and dropped my constants of integration and absolute values but the result is the same. So the question is why is the integral of 1/x ln(x), or why is 1/x the derivative of the natural log function?

y = ln(x)
e^y = x
dy/dx e^y = 1
dy/dx x = 1
dy/dx = 1/x

There you go

This post was edited by nobrow on Jun 21 2015 12:03am

And what about f(x) = 2.exp (x) ?

Actually exponential is the sole function y such that :
y ' = y
y(0) = 1

Oh yea.....forgot about that initial condition. Details details....

Just make your own proof using your limit definition and this becomes clear.

Let f(x)=e^x

Then lim h->0 of f(x+h)-f(x)/h
=lim h->0 of e^(x+h)-e^x/h
=lim h->0 e^h*e^x-e^x/h
=e^x lim h->0 e^h-1/h
=e^x

Thus f(x)=f'(x) when f(x)=e^x

This post was edited by Xx Shin3d0wn xX on Jun 22 2015 09:13am

Casey pls.

e = 2.718281828... (I like how it does the 18 28 twice, which is the only reason i give two shits about e's value).

But srsly.

Casey pls.

Casey pls.

math is so fun