Solving for derivative of e^x using limit definition

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  • #1
dnt
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Homework Statement



I want to solve for the derivative of e^x using the limit definition.

Homework Equations



http://www.math.hmc.edu/calculus/tutorials/limit_definition/img10.png [Broken]

The Attempt at a Solution



obviously the derivative of e^x is itself, so i konw the answer. i just cannot derive it using the limit definition.

i plug it in and i get lim (h->0) [ e^(x+h) - e^x ] / h

with h being delta x

from there i have no idea how to simplify it. can someone give me a hint on how to start and some possibly ways to simplify it down?

thanks.
 
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Answers and Replies

  • #2
Hint:

You can simplify what you have.

[tex]e^{x+h}[/tex] can be rewritten.

[Edit 1]
Ooooo... sneaky. Even with this, you'll still need to evaluate the limit, and both L'hopital and Taylor expansions require a priori knowledge of the derivative...

[Edit 2]
Ok, someone better at real analysis would have to double check me on this, but under appropriate "nice" conditions, you should be able to transform from

[tex]\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}[/tex]
to
[tex]\lim_{g(h)\rightarrow0}\frac{f(x+g(h))-f(x)}{g(h)}[/tex]

I would choose [tex]g(h)=\hbox{ln}(h)[/tex]

ZM
 
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  • #3
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d(f^x)/dx=[f^(x+d)-f^x]/d = f^x[f^d-1]/d

Let's us assume that there exists a number, e such that e^d=1+d as d gets small.

You can find this number from the limit d->0 of

e=(1+d)^(1/d)

A calculator shows that for e.g. d=1/1000, e=2.7169
The limit goes to e=2.71828182846

Finally, substitution of f=e gives

d(e^x)/dx = e^x[e^d-1]/d = e^x

(using the fact that e obeys e^d=1+d, for small d)
 
  • #4
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Homework Statement



I want to solve for the derivative of e^x using the limit definition.
obviously the derivative of e^x is itself, so i konw the answer. i just cannot derive it using the limit definition.

i plug it in and i get lim (h->0) [ e^(x+h) - e^x ] / h
You have:
[tex]\frac{d}{dx} \, e^x = \lim_{h \rightarrow 0} \frac{e^{x + h} - e^x}{h} = \lim_{h \rightarrow 0} \frac{e^x(e^h - 1)}{h}[/tex]

so all you need to show is that:

[tex]\lim_{h \rightarrow 0} \frac{e^h - 1}{h} = 1[/tex]

which you can do by using the definition of limit.
 
  • #5
arildno
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Really?
What you actually need is a proper definition of what it means to raise a number to the power of a real number.
 
  • #6
You have:
[tex]\frac{d}{dx} \, e^x = \lim_{h \rightarrow 0} \frac{e^{x + h} - e^x}{h} = \lim_{h \rightarrow 0} \frac{e^x(e^h - 1)}{h}[/tex]

so all you need to show is that:

[tex]\lim_{h \rightarrow 0} \frac{e^h - 1}{h} = 1[/tex]

which you can do by using the definition of limit.
Just out of curiosity, how do you do that? Note that, when you set [tex]h=0[/tex] you find that the limit is equal to [tex]\frac{0}{0}[/tex] which is undefined. So you either need to use L'Hopital's rule---which requires knowing the derivative of [tex]e^x[/tex] or you need to use a Taylor expansion---which requires knowing the derivative of [tex]e^x[/tex].

Of course, you can't use the derivative of [tex]e^x[/tex] because this is what you're solving for.... Hence my "sneaky" comment above.

I think the only way to run is to use the derivative of the natural log function as I defined above. But someone better at real analysis would have to go over my work.

ZM
 
  • #7
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Didn't I show this in my post?
 
  • #8
arildno
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ONLY way, dear?
Nope!

We just define a function called Exp(x) as follows:
[tex]Exp(x)=1+\sum_{n=1}^{\infty}\frac{x^{n}}{n!}[/tex]
We can prove that this is a very nice invertible function, and if we call the inverse Log(x), we may define general power functions as:
[tex]a^{x}\equiv{E}xp(x*Log(a)), a>0[/tex]
In particular, if we define a number "e" as e=Exp(1), we gain the results we want.

In particular, this series can be proven to have all the features we like the exponential function to have, without regarding it as the "Taylor expansion of e^x".
 
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  • #9
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Just out of curiosity, how do you do that? Note that, when you set [tex]h=0[/tex] you ...
When finding the limit, you don't just set h = 0. That is very bad! On second thought, the definition of limit does not help in finding the limit of a function. I retract what I said about that.

christianjb said that:

[tex]e^x = 1 + x[/tex]

for small x. Where did you get this fact from christianjb?
 
  • #10
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When finding the limit, you don't just set h = 0. That is very bad! On second thought, the definition of limit does not help in finding the limit of a function. I retract what I said about that.

christianjb said that:

[tex]e^x = 1 + x[/tex]

for small x. Where did you get this fact from christianjb?
Here's what I wrote:
"Let's us assume that there exists a number, e such that e^d=1+d as d gets small.

You can find this number from the limit d->0 of

e=(1+d)^(1/d)

A calculator shows that for e.g. d=1/1000, e=2.7169
The limit goes to e=2.71828182846"


I didn't prove that (1+d)^(1/d) converges, but if it does converge- then e^d=1+d for small values of d, (by defn. of e).
 
  • #11
ONLY way, dear?
Nope!

We just define a function called Exp(x) as follows:
[tex]Exp(x)=1+\sum_{n=1}^{\infty}\frac{x^{n}}{n!}[/tex]
We can prove that this is a very nice invertible function, and if we call the inverse Log(x), we may define general power functions as:
[tex]a^{x}\equiv{E}xp(x*Log(a)), a>0[/tex]
In particular, if we define a number "e" as e=Exp(1), we gain the results we want.
This is well and good, but appears ad-hoc. We started with a known function, but wound up needing to know the derivative of this function to calculate the derivative of this function.

Now, if you can play with your fingers and toes for a while and come up with the infinite series which can be shown to converge to [tex]e^x[/tex] (without, of course, using the derivative of this function---because that's what we're looking for), then you're a better mathematician than I (not hard, actually).

The only way out of this one is to begin by defining [tex]e^x[/tex] as the infinite series you TeXed so well above. Then everything should follow as long as one shows convergence.

In particular, this series can be proven to have all the features we like the exponential function to have, without regarding it as the "Taylor polynomial of e^x".
Again, I just want to be sure there's no circular logic. You can't use the derivative of a function to define the derivative of that very same function.

ZM
 
  • #12
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This is well and good, but appears ad-hoc. We started with a known function, but wound up needing to know the derivative of this function to calculate the derivative of this function.

Now, if you can play with your fingers and toes for a while and come up with the infinite series which can be shown to converge to [tex]e^x[/tex] (without, of course, using the derivative of this function---because that's what we're looking for), then you're a better mathematician than I (not hard, actually).

The only way out of this one is to begin by defining [tex]e^x[/tex] as the infinite series you TeXed so well above. Then everything should follow as long as one shows convergence.



Again, I just want to be sure there's no circular logic. You can't use the derivative of a function to define the derivative of that very same function.

ZM
I disagree. It's just a defn. of e. You have to define e somehow. I gave one definition- and this is another. Neither of our definitions involve a Taylor series expansion. The above is equivalent to a Taylor series, but that's neither here nor there.

I think the only problem is showing that the limit exists.
 
  • #13
I disagree. It's just a defn. of e. You have to define e somehow. I gave one definition- and this is another.
Like I said, if you start with that definition, then you're golden. You just can't say "here's a function [tex]e^x[/tex] of which we know some properties. Oh, wait, but now we need to use the derivative of this function to calculate the derivative of this function. So let's posit that this function can be represented by this infinite series..."

...unless you've already proven that the series has the requisite correspondence.

I think the only problem is showing that the limit exists.
I think we're all agreeing here... just niggling on details....

ZM
 
  • #14
arildno
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The only way out of this one is to begin by defining [tex]e^x[/tex] as the infinite series you TeXed so well above. Then everything should follow as long as one shows convergence.
Why can I not define e=Exp(1)?

The series, as stated, can AS A FUNCTION ITSELF, be proven to have those properties we would like an exponential to have.
In particular, for any open R-disk about x=0, the sequence of partial sum functions is uniformly convergent.
Thus, by using the Cauchy product of infinite series, you may prove the exponential property Exp(x+y)=Exp(x)*Exp(y).

The limit of the differentiated partial sums is also convergent, which is crucial in proving that Exp(x) is differentiable, having itself as its derivative.

This again can be used to prove that Exp(x) is invertible.

Note that I have used powers of NATURAL numbers in the series expansion.
This has been done, since these powers can be defined independently, by induction. (It can be proven that using x as a natural number in the a^x formulation will agree with the results from the inductive definition of these powers)


Again, I just want to be sure there's no circular logic.
there is none.
You can't use the derivative of a function to define the derivative of that very same function.
Indeed you can't.
But it is perfectly permissible to filch ideas from somewhere else in order to build another, wholly rigorous proof.
Where ideas are gotten from is not part of the proofs, only the stated&sufficient presuppositions&axioms.
 
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  • #15
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Like I said, if you start with that definition, then you're golden. You just can't say "here's a function [tex]e^x[/tex] of which we know some properties. Oh, wait, but now we need to use the derivative of this function to calculate the derivative of this function. So let's posit that this function can be represented by this infinite series..."

...unless you've already proven that the series has the requisite correspondence.



I think we're all agreeing here... just niggling on details....

ZM
I agree- we all agree. However, my car's still faster than yours.
 
  • #16
dextercioby
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You have:
[tex]\frac{d}{dx} \, e^x = \lim_{h \rightarrow 0} \frac{e^{x + h} - e^x}{h} = \lim_{h \rightarrow 0} \frac{e^x(e^h - 1)}{h}[/tex]

so all you need to show is that:

[tex]\lim_{h \rightarrow 0} \frac{e^h - 1}{h} = 1[/tex]

which you can do by using the definition of limit.
Actually it's the definition of "e" and the fact that the limit to infinity and the natural log function commute. Here's the trick. Assume you have the function [itex] f(x) [/itex] which is defined on [itex] (0,\infty) [/itex] and is never zero on its domain. Hence [itex] 1/f(x) [/itex] is an equally well defined function on the domain of "f", namely [itex] (0,\infty) [/itex].

Assume that [itex] \lim_{x\rightarrow 0} f(x)=a [/itex], where "a" is any number greater than 0 and smaller than infinity. Hence a^{-1} is a number having the same property. The function "f" is obviously assumed to be continuous on all of its domain. Now, after all these assumptions, we're in position to infer that

[tex] \lim_{x\rightarrow 0}\frac{1}{f(x)}=\frac{1}{\lim_{x\rightarrow 0} f(x)}=\frac{1}{a} [/tex].

Let's apply the above considerations to our case.

[tex] \lim_{x\rightarrow 0}\frac{e^{x}-1}{x} \ = \ ... ?[/tex]

Let's choose [itex] f(x)=\frac{x}{e^{x}-1} [/itex]. It satisfies all the domain, continuity and valuedness issues spelled above. All we have to do is compute [itex] \lim_{x\rightarrow 0} f(x) [/itex] and show that the result is a "good number", namely any number but 0 and also the result shouldn't be "+infinity".

[tex] \lim_{x\rightarrow 0}\frac{x}{e^{x}-1} = \lim_{h\rightarrow 1}\frac{\ln h}{h-1} [/tex] , where i used the substitution x=ln h

[tex] =\lim_{\bar{h}\rightarrow 0}\frac{\ln (\bar{h}+1)}{\bar{h}} [/tex] , where i used the substitution \bar{h}=h-1

[tex] =\lim_{\bar{h}\rightarrow 0}\ln (\bar{h}+1)^{\frac{1}{\bar{h}}} =\ln \lim_{\bar{h}\rightarrow 0}(\bar{h}+1)^{\frac{1}{\bar{h}}} =\ln e =1 [/tex], where i used the definition of "e".

Therefore i obtained that [itex] \lim_{x\rightarrow{0}}f(x)=1 [/itex] and i can infer that [itex] \lim_{x\rightarrow{0}}\frac{1}{f(x)}= \lim_{x\rightarrow{0}}\frac{e^x -1}{x}=1 [/itex].
 
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  • #17
HallsofIvy
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Defining ex as
[tex]e^x= \sum_{n=0}^\infty \frac{x^n}{n!}
\text{ is perfectly valid. It's just different from the definition you want to use.}\\
\\ \text{It is also perfectly valid to define }
\ln(x)= \int_1^x \frac{1}{t}dt\ ,[/tex]
show that it is one-to-one and then define ex as its inverse function. That makes the derivative of both ln(x) and ex easy.
 
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  • #18
Gib Z
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There are 4 basic definitions of e from what I can remember, and most others can be easily seen from these 4.

1.[tex]e^x=\lim_{n\to\infty} (1+ \frac{x}{n})^n[/tex]

2. [tex]e^x= \sum_{n=0}^\infty \frac{x^n}{n!}[/tex]

3. The unique real number such that [tex]\int_1^e \frac{1}{t} dt = 1[/tex]

4. [tex]\frac{d}{dx} e^x = e^x[/tex].

Now what you could have done is defined e to be definition 2, 3 or 4, and shown how that is equal to the limit definition.

Heres a link that does that for you : http://en.wikipedia.org/wiki/Charac...function#Equivalence_of_the_characterizations
 
  • #19
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This is for all the first year calc students who found this thread looking for an acceptable proof.

y=e^x

take ln of both sides

lny = xlne

differentiate implicitly and use limit definition

(1/y)y' = limx-0 [(x+h)lne - xlne]/h

factor out lne

(1/y)y' = [(lne) x + h -x]/h

(1/y)y' = (lne)h/h
(1/y)y' = (lne)

multiply both sides by y

y' = lne(y)

but y=e^x

y' = e^x lne

but lne is= 1

y'=e^x

Note: you can repeat this with any base, using FIRST YEAR intro calc one.
the last step would set y' = b^x lnb
 
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  • #20
hunt_mat
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Can't you just use that:
[tex]
e^{x}=\lim_{n\rightarrow\infty}\left( 1+\frac{x}{n}\right)^{n}
[/tex]
Differentiate that using the chain rule to obtain:
[tex]
(e^{x})'(x)=\lim_{n\rightarrow\infty}\left( 1+\frac{x}{n}\right)^{n-1}
[/tex]
and then write:
[tex]\left( 1+\frac{x}{n}\right)^{n-1}=\frac{\left( 1+\frac{x}{n}\right)^{n}}{1+\frac{x}{n}}
[/tex]
and then take the limit as [itex]n\rightarrow\infty[/itex] and use the algebra of limits?
 
  • #21
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Why not just use L'Hopital's Rule in post #4 on


Limit as h→0 e^h − 1 / h
 
  • #22
Another way to do it is to instead show that the derivative of ln x is 1/x using a limit of a difference quotient and substituting the limit defintion of e^x which was the first definition given:

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

So you can show that the derivative of ln x is 1/x purely by using the above defintion of e^x. After you have done that, you can show that the derivative of the inverse function of ln x is 1/(1/(e^x)) = e^x. Note that the entire proof rests on the limit defintion of e^x and the fact that ln x is the inverse function of e^x.

I also want to mention that the proofs laid out by some of our esteemed members are not 'circular logic'. At the end of the day you have to define something and build on top of that. You can define e^x to be the limit above. You can define e^x to be the Taylor series. You can even avoid defining e^x directly and instead define the integral of ln x and then define e^x relative to that but you have to start somewhere. As such, the approaches are valid. I think the reason the original poster may find this approach more appealing is that the limit definition of e^x (mentioned above) is not as readily obvious from the fact that the derivative of e^x is e^x while the other definitions 'seem' outwardly to be a consequence of the derivative of e^x being e^x.

All the best,
Junaid Mansuri
 
  • #23
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You have:
[tex]\frac{d}{dx} \, e^x = \lim_{h \rightarrow 0} \frac{e^{x + h} - e^x}{h} = \lim_{h \rightarrow 0} \frac{e^x(e^h - 1)}{h}[/tex]

so all you need to show is that:

[tex]\lim_{h \rightarrow 0} \frac{e^h - 1}{h} = 1[/tex]

which you can do by using the definition of limit.
I just want to point out one way of using this method to solve this problem since it seems to be non-obvious to some people.

Write e^h as a series 1+ h + h^2/2 +...

[tex]\lim_{h \rightarrow 0} \frac{e^x(e^h - 1)}{h} = \lim_{h \rightarrow 0} e^x \frac{ 1 + h + h^2/2+ ... - 1} {h} = \lim_{h \rightarrow 0} e^x(1 + h/2 + h^2 / 6 + ...) = e^x[/tex]
 

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