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Solving for derivative of e^x using limit definition

  1. Mar 22, 2007 #1


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    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

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

    3. 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?

    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Mar 22, 2007 #2

    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


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

    Last edited: Mar 22, 2007
  4. Mar 22, 2007 #3
    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


    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)
  5. Mar 22, 2007 #4
    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.
  6. Mar 22, 2007 #5


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    What you actually need is a proper definition of what it means to raise a number to the power of a real number.
  7. Mar 22, 2007 #6
    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.

  8. Mar 22, 2007 #7
    Didn't I show this in my post?
  9. Mar 22, 2007 #8


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    ONLY way, dear?

    We just define a function called Exp(x) as follows:
    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".
    Last edited: Mar 22, 2007
  10. Mar 22, 2007 #9
    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?
  11. Mar 22, 2007 #10
    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


    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).
  12. Mar 22, 2007 #11
    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.

  13. Mar 22, 2007 #12
    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.
  14. Mar 22, 2007 #13
    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....

  15. Mar 22, 2007 #14


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    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)

    there is none.
    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.
    Last edited: Mar 22, 2007
  16. Mar 22, 2007 #15
    I agree- we all agree. However, my car's still faster than yours.
  17. Mar 23, 2007 #16


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    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].
    Last edited: Mar 23, 2007
  18. Mar 23, 2007 #17


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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.
    Last edited by a moderator: Aug 10, 2013
  19. Mar 23, 2007 #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
  20. Dec 11, 2011 #19
    This is for all the first year calc students who found this thread looking for an acceptable proof.


    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


    Note: you can repeat this with any base, using FIRST YEAR intro calc one.
    the last step would set y' = b^x lnb
    Last edited: Dec 11, 2011
  21. Dec 11, 2011 #20


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    Can't you just use that:
    e^{x}=\lim_{n\rightarrow\infty}\left( 1+\frac{x}{n}\right)^{n}
    Differentiate that using the chain rule to obtain:
    (e^{x})'(x)=\lim_{n\rightarrow\infty}\left( 1+\frac{x}{n}\right)^{n-1}
    and then write:
    [tex]\left( 1+\frac{x}{n}\right)^{n-1}=\frac{\left( 1+\frac{x}{n}\right)^{n}}{1+\frac{x}{n}}
    and then take the limit as [itex]n\rightarrow\infty[/itex] and use the algebra of limits?
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