Differentiation of e^x from first principles

So I was just messing around with stuff and I'm a bit confused. I heard that the derivative of e^x was e^x, so I tried to differentiate from first principles.

I'm left with:

\displaystyle e^x\lim_{h\to0} \frac {(e^h-1)}{h}

Which I think is right.

But if I let h tend to 0, then don't I get

\frac {0}{0}

, which is undefined? Is that it then? :confused:

Reply 1

SR255

What is the limit definition of the derivative of f(x)? Plug e^x into that and see what you get.

Reply 2

such

5

You're right so far. However, the answer is not undefined. Hint: try using the Maclaurin expansion of e^h.

(edited 10 years ago)

Reply 3

Big-Daddy

13

I thought e was defined such that the differential of e^x with respect to x is e^x? In which case trying to prove this is useless ...

Reply 4

Smaug123

15

Original post by Big-Daddy

I thought e was defined such that the differential of e^x with respect to x is e^x? In which case trying to prove this is useless ...

It depends where you start from. I think the usual definition (certainly the one I learnt) is

\displaystyle \sum_{i=0}^{\infty} \dfrac{x^i}{i!}

.

Reply 5

Smaug123

15

Original post by ThatPerson

\displaystyle e^x\lim_{h\to0} \frac {(e^h-1)}{h}

The limit you have left to consider is very closely related to the derivative of

e^x

. Can you see how?

Reply 6

ThatPerson

OP

18

Original post by Smaug123

The limit you have left to consider is very closely related to the derivative of

e^x

. Can you see how?

I think I need to go further into this topic and then come back to this, because some of this stuff has gone straight over my head.

brb in 6 weeks :tongue:

Reply 7

brianeverit

9

Original post by Big-Daddy

I thought e was defined such that the differential of e^x with respect to x is e^x? In which case trying to prove this is useless ...

I have always thought that the definition of e is

e=\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^n

Reply 8

Intriguing Alias

16

Original post by brianeverit

I have always thought that the definition of e is

e=\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^n

Certainly in analysis we began with a power series definition and derived all other properties from that. If you cannot derive all other properties using your 'definition' then it isn't one.

Reply 9

bananarama2

Original post by Intriguing Alias

Certainly in analysis we began with a power series definition and derived all other properties from that. If you cannot derive all other properties using your 'definition' then it isn't one.

But it was the original definition, so surely all the properties can be derived?

Reply 10

Intriguing Alias

16

Original post by bananarama2

But it was the original definition, so surely all the properties can be derived?

I wasn't saying it wasn't - sorry that was just a more general comment. I'm personally not sure whether certain definitions are valid or not :smile:

But to comment on that specifically - that'd be a definition of e but doesn't define e^x unless you assume the existence of exponential functions. That's a bit of a different issue, though, and isn't very related to the depth of this thread.

(edited 10 years ago)

Reply 11

moritzplatz

12

Original post by bananarama2

But it was the original definition, so surely all the properties can be derived?

surely that's the definition of e.

how do you define powers (non integer ones) then though?

Reply 12

Smaug123

15

Original post by bananarama2

But it was the original definition, so surely all the properties can be derived?

All these definitions are equivalent to each other. It takes differing amounts of work to prove this from each definition - the power series one is the easiest to prove things about.

Reply 13

Smaug123

15

Original post by moritzplatz

surely that's the definition of e.

how do you define powers (non integer ones) then though?

x^a \equiv \exp(\log(x) \times a)

.

Reply 14

moritzplatz

12

Original post by Smaug123

x^a \equiv \exp(\log(x) \times a)

.

ahah
you see you get back to the orginal problem.
how do you define exp.

p.s. I was just pointing out to the previous poster why his definition was not complete

it would be fine replacing the suitable 1 by x in the limit though

Reply 15

bananarama2

Original post by bananarama2

But it was the original definition, so surely all the properties can be derived?

Original post by Intriguing Alias

I wasn't saying it wasn't - sorry that was just a more general comment. I'm personally not sure whether certain definitions are valid or not :smile:

But to comment on that specifically - that'd be a definition of e but doesn't define e^x unless you assume the existence of exponential functions. That's a bit of a different issue, though, and isn't very related to the depth of this thread.

Original post by moritzplatz

surely that's the definition of e.

how do you define powers (non integer ones) then though?

There must be a way. I just haven't found it yet :tongue:

Reply 16

bananarama2

Original post by Smaug123

All these definitions are equivalent to each other. It takes differing amounts of work to prove this from each definition - the power series one is the easiest to prove things about.

From a personal point of view, whenever I prove something I like to do it from a sort of historical background. I like understanding why stuff is what it is. That's a personal thing though.

Reply 17

Smaug123

15

Original post by moritzplatz

ahah
you see you get back to the orginal problem.
how do you define exp.

\displaystyle \exp(x) \equiv \sum_{i=0}^{\infty} \dfrac{x^i}{i!}

.

Original post by bananarama2

From a personal point of view, whenever I prove something I like to do it from a sort of historical background. I like understanding why stuff is what it is. That's a personal thing though.