Is derivative of e^x defined or can it be proved...

A

apple2357

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Ok, i am going around in circles here...

How do you prove that the derivative of e^x is e^x?

If you do it by first principles, you need a result ( namely the lim of (e^h-1)./h as h tends to zero is 1) and this result appears to depend on what we are trying to prove? Unless anyone can offer me another way of thinking about this limit result.

So is it a definition?
 
apple2357 said:
Ok, i am going around in circles here...

How do you prove that the derivative of e^x is e^x?

If you do it by first principles, you need a result ( namely the lim of (e^h-1)./h as h tends to zero is 1) and this result appears to depend on what we are trying to prove? Unless anyone can offer me another way of thinking about this limit result.

So is it a definition?
Click to expand...
If it were just a definition, then you'd have to prove that it is consistent with the definition of the derivative itself.

So, yes, you can prove that the derivative of e^x, according to the definition of the derivative, is e^x.

To do that, you have to use the definition of e^x! So, what is that definition?
 
that sounds circular to me!
So i could define e^x as a infinite series?

But can i not just simply define a function which differentiates to itself and call it e^x and then there is no need to prove anything?
And if that is defined, i can then go on to look at the implications? One implication being the (e^h-1)=1 as h tends to zero.

I guess i am getting at.. what is the universally accepted starting point for e^x within the mathematical community?
 
apple2357 said:
that sounds circular to me!
So i could define e^x as a infinite series?

But can i not just simply define a function which differentiates to itself and call it e^x and then there is no need to prove anything?
And if that is defined, i can then go on to look at the implications? One implication being the (e^h-1)=1 as h tends to zero.

I guess i am getting at.. what is the universally accepted starting point for e^x within the mathematical community?
Click to expand...
How is it circular to work back from one definition to another? Unless, of course, you have defined e^x in terms of a derivative -- but that's why I asked! One way would, indeed, be to define e as a series.

Yes, you can define e^x as the solution of a simple differential equation; if you do so, then you have to prove other facts about it, such as the value of e^1, and its properties (ultimately, the fact that it is an exponential function at all).

There are many possible starting points, and in different textbooks, you can find different routes through all of this. The important thing is that, wherever you start, you'll get all the same information.

Have you tried looking this up? The Wikipedia article on the exponential function has a link here:

Characterizations of the exponential function - Wikipedia

en.wikipedia.org en.wikipedia.org

That shows six different starting points, and proofs that they give equivalent results. Characterization #4 is your suggestion.
 
Assuming that you accept what the derivative of ln(x) is 1/x, then just use x = ln(e^x). Now if take the derivative you will find out what the derivative of e^x equals.
 
Thank you Dr P, that link is exactly what i was looking for. I didn't realise there were so many potential starting points. i had assumed there was an agreed one.
 
Steven G said:
Assuming that you accept what the derivative of ln(x) is 1/x, then just use x = ln(e^x). Now if take the derivative you will find out what the derivative of e^x equals.
Click to expand...
x= ln(e^x)
(x)' = (ln(e^x))'
1 = (1/e^x)(e^x)'
Therefore, (e^x)' = e^x
 
apple2357 said:
Thank you Dr P, that link is exactly what i was looking for. I didn't realise there were so many potential starting points. i had assumed there was an agreed one.
Click to expand...
First, they are all equivalent.

My recollection is that the series whose limit equals the definite integral for the hyperbola was the historical starting point (that is basically where Steven is pointing you).
 
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