Derivative of e

[DJBruce]

I have been trying to teach myself calculus and I came upon this:

[math]\frac{d}{dx}e^{x}=e^{x}[/math]

I have been struggling to understand why this is. From my basic knowledge I have got this far.

[math]\frac{d}{dx}e^{x}=\frac{e^{x+h}-e^{x}}{(x+h)-x}[/math]

[math]\frac{d}{dx}e^{x}=\stackrel{Lim}h{\rightarrow}0\frac{e^{x+h}-e^{x}}{(x+h)-x}=\frac{e^{x}-e^{x}}{(x)-x}=\frac{0}{0}[/math]

So I keep coming up with this intermediate form and I cannot figure out how to get rid of it. I might be doing this completely wrong so could some one please point me in the right direction.

[ecoli]

here's the proof: http://math2.org/math/derivatives/more/e%5Ex.htm

[Kyrisch]

I have been trying to teach myself calculus and I came upon this:

[math]\frac{d}{dx}e^{x}=e^{x}[/math]

I have been struggling to understand why this is. From my basic knowledge I have got this far.

[math]\frac{d}{dx}e^{x}=\frac{e^{x+h}-e^{x}}{(x+h)-x}[/math]

[math]\frac{d}{dx}e^{x}=\stackrel{Lim}h{\rightarrow}0\frac{e^{x+h}-e^{x}}{(x+h)-x}=\frac{e^{x}-e^{x}}{(x)-x}=\frac{0}{0}[/math]

So I keep coming up with this intermediate form and I cannot figure out how to get rid of it. I might be doing this completely wrong so could some one please point me in the right direction.

 

That limit is just a definition of the derivative. It's really only useful in actually solving for the derivative algebraically with polynomial functions. With most others, you end up with indeterminate forms.

[Dave]

I have been trying to teach myself calculus and I came upon this:

[math]\frac{d}{dx}e^{x}=e^{x}[/math]

I have been struggling to understand why this is. From my basic knowledge I have got this far.

[math]\frac{d}{dx}e^{x}=\frac{e^{x+h}-e^{x}}{(x+h)-x}[/math]

[math]\frac{d}{dx}e^{x}=\stackrel{Lim}h{\rightarrow}0\frac{e^{x+h}-e^{x}}{(x+h)-x}=\frac{e^{x}-e^{x}}{(x)-x}=\frac{0}{0}[/math]

So I keep coming up with this intermediate form and I cannot figure out how to get rid of it. I might be doing this completely wrong so could some one please point me in the right direction.

 

Your logic is not correct; specifically, the denominator of your limit is incorrect. It is, in fact, perfectly possible to obtain the derivative through the limit definition. Let [imath]f(x)=e^x[/imath], then we have that:

 

[math]f'(x) = \lim_{h\to 0} \frac{e^{x+h} - e^x}{(x+h) - x} = \lim_{h\to 0} \frac{e^x(e^h - 1)}{h} = e^x \lim_{h\to 0} \frac{e^h - 1}{h}[/math]

 

Now, if you accept that I can write [imath]e^x[/imath] in the common Taylor series form:

 

[math]e^{h} = 1 + h + \frac{h^2}{2!} + \frac{h^3}{3!} + \cdots = \sum_{k=0}^{\infty} \frac{h^k}{k!}[/math]

 

then

 

[math]e^{h} - 1 = h + \frac{h^2}{2!} + \frac{h^3}{3!} + \cdots = \sum_{k=1}^{\infty} \frac{h^k}{k!}[/math]

 

and so

 

[math]\frac{e^{h} - 1}{h} = 1 + \frac{h}{2!} + \frac{h^2}{3!} + \cdots = \sum_{k=0}^{\infty} \frac{h^k}{(k+1)!}[/math]

 

giving

 

[math]\lim_{h\to 0} \frac{e^{h} - 1}{h} = 1 + 0 + 0 + \cdots = 1[/math]

 

Hence,

 

[math]f'(x) = e^x \cdot 1 = e^x[/math].

[the tree]

Now, if you accept that I can write [imath]e^x[/imath] in the common Taylor series form:

If you're going to do that, then wouldn't it be simpler just to look at the derivative of each term in the series and that it is the previous term?

[Dave]

If you're going to do that, then wouldn't it be simpler just to look at the derivative of each term in the series and that it is the previous term?

 

No, because you would have to prove that you can do such a thing. It is not trivial that

 

[math]f(x) = \sum_{n=0}^{\infty} a_n x^n \Rightarrow f'(x) = \sum_{n=1}^\infty n a_n x^{n-1}[/math]

[the tree]

Of course. My objection was that the Taylor series is dependant on knowing the derivative. Depending on how you define e^x then I think the proof that e-coli linked to is basically sufficient.

[Dave]

Of course. My objection was that the Taylor series is dependant on knowing the derivative. Depending on how you define e^x then I think the proof that e-coli linked to is basically sufficient.

 

Sure, the proof in ecoli's link is sufficient (if a little non-rigourous) but it's not the answer to the original question, which was to calculate the derivative using limits.

 

In this case, I believe that if you want to do this problem properly then you should start from the ground up and use the minimal amount of mathematics possible. (If you didn't want to do this, then why would you be considering this specific problem in the first place?) That means defining the function properly; most commonly, we can do that with a limit form, such as

 

[math]e^x = \lim_{n\to\infty} \left( 1 + \frac{x}{n} \right)^n[/math]

 

which you can show to be equivalent to the power series with a little work. This enables you can to calculate the derivative knowing very little at all about power series and nothing at all about logarithms.

[mooeypoo]

btw, whenever you have a limit that ends with 0/0 or inf/inf, you need to simplify it (use L'hopital rule or anything else to continue). 0/0 for limit is meaningless.

 

Or.. uhm.. correct me if I'm wrong? that's the way I studied it, and that's what jumped right away when I first saw the post.

 

So I'm not quite sure how I would continue this limit, but I wouldn't settle if it ended like that..

[Dave]

btw, whenever you have a limit that ends with 0/0 or inf/inf, you need to simplify it (use L'hopital rule or anything else to continue). 0/0 for limit is meaningless.

 

Or.. uhm.. correct me if I'm wrong? that's the way I studied it, and that's what jumped right away when I first saw the post.

 

So I'm not quite sure how I would continue this limit, but I wouldn't settle if it ended like that..

 

That's what I've done?

[Harlequinne]

e is a number spiral written on squared paper.1. up to two.one to the right to three.one down to four and again one down to five.THAT relationship is what e is.that expanse of a square...the same unending square.so it also has no rate of change decipherable and the area underneath is the same ratio.

[Dave]

Okay... or not.

[the tree]

... that didn't make much sense, did it?

[Dave]

No, not really