Talk:Derivative of Exponential Function

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$\frac{dy}{dx}=\frac{e^x(e^h-1)}{h}$

$\lim_{h \to 0}{(e^h-1)}=h$

$\frac{dy}{dx}=\frac{e^xh}{h}$

Is the second line above correct? It seems to me that as h → 0, eh→ 1 and (eh --1)→ 0

Definition of e

According to Stewart's Calculus: Early Transcendentals, the definition of the number e is the number such that its derivative at x=0 is 1.

This is equivalent to stating $~\lim_{h \to 0} \frac{e^h - 1}{h} = 1 $.

Why? This is simply because

$f(x)=e^{x}\rightarrow f'(x)=\lim_{h \to 0}\frac{e^{x+h}-e^{x}}{h} = \lim_{h \to 0}\frac{e^{x}(e^{h} - 1)}{h}$

Defining $f'(0)=1$, we get

$ 1 = f'(0) = \lim_{h \to 0}\frac{e^{0}(e^{h} - 1)}{h} = \lim_{h \to 0}\frac{1\cdot(e^{h}-1)}{h}=\lim_{h \to 0}\frac{e^{h}-1}{h}$

There ya go.

-- syntax fixed by Paul E J King (talk) 23:42, 27 April 2013 (UTC)

Using e to define e ... ?

I always taught that e was defined as: $e = \lim_{n \to \infty} (1+\frac{1}{n})^n$ Paul E J King (talk) 09:18, 30 April 2013 (UTC)

There are several ways to define $e$. One of them is the limit of that sequence. There are others. They can be found in the (currently undergoing refactoring) page Definition:Euler's Number. --prime mover (talk) 10:11, 30 April 2013 (UTC)

Infinite Series

Take a common definition of the exponential function: $e^x = \sum_{n = 0}^{\infty} {\frac{x^n}{n!}}$

Evaluating the limit based on this definition: $\lim_{h \to 0} \frac{e^h - 1}{h}=\lim_{h \to 0} \frac{\sum_{n = 0}^{\infty} {\frac{h^n}{n!}}-1}{h}$

$=\lim_{h \to 0} \frac{\frac{h^0}{0!}+\sum_{n = 1}^{\infty} {\frac{h^n}{n!}}-1}{h}$

$=\lim_{h \to 0} \frac{1+\sum_{n = 1}^{\infty} {\frac{h^n}{n!}}-1}{h}$

$=\lim_{h \to 0} \sum_{n = 1}^{\infty} {\frac{h^{n-1}}{n!}}$

$=\lim_{h \to 0} \sum_{n = 0}^{\infty} {\frac{h^{n}}{(n+1)!}}$

$=\sum_{n = 0}^{\infty} {\frac{0^n}{(n+1)!}}$

$=\frac{0^0}{(0+1)!}+\sum_{n = 1}^{\infty} {\frac{0^n}{(n+1)!}}$

$=\frac{1}{1}+0$

$=1$

Notation

I find that $e^x$ is a lot easier to read than $\exp x$. Of course, that could just be me, or it could be that the exp notation is more common in higher mathematics. --Cynic (talk) 22:21, 22 January 2009 (UTC)

The operation of associating $2.718281828...^x \ $ is a multifunction in $\C \ $, whereas the map $\text{ exp }:\C \to \C$ is a well-defined transformation of the complex plane that corresponds to the principal branch of $2.718282828...^x \ $. Frequently, the exponential function is simply abbreviated $e^x \ $, with the understanding that it refers only to exponential function and not to $e=2.718281828...., \ $ raised to the $x \ $ power. I haven't contributed to anything regarding exponentials too heavily, but that's a necessary distinction when discussing the exponential as a complex function, though unnecessary and cumbersome for real analysis or any calculus at an elementary level. Zelmerszoetrop 22:29, 22 January 2009 (UTC)

As you say, the distinction is necessary at complex analysis level (which I'm working towards, I'll start as soon as I've finished up defining the trig functions in terms of $D_{xx} f \left({x}\right) = -A f \left({x}\right)$.

First proof

Hello, the argument in the first proof for the claim that (e^h - 1)/h --> 1 as h --> 0 was circular (it had used l'Hopital's rule, which already assumed that the derivative of e^x is e^x). I changed it to a new one; I hope someone will point out if I've made an error. Mag487 23:59, 22 August 2009 (UTC)

What part of l'hopital's rule depended on d(e^x)/dx = e^x? I'm probably being dense here... --Cynic (talk) 02:59, 23 August 2009 (UTC)

Sorry, I was unclear. The proof relied (it has referred the reader to this page) on using l'Hopital's rule on the quantity $\lim_{h\to 0} \frac{\exp h - 1}{h}$ to show that it's equal to one. However, in order to apply the rule here, we have to already know the derivative of the function e^x in the numerator. Mag487 05:07, 23 August 2009 (UTC)

Ah, I see. Thanks a lot for fixing it! --Cynic (talk) 00:40, 24 August 2009 (UTC)