Is there a way to prove that the derivative of $e^x$ is $e^x$ without using chain rule? If so, what is it? Thanks.
Proof of derivative of $e^x$ is $e^x$ without using chain rule
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0do you know about Taylor series? – 2012-09-20
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0Taylor series assumes the existence of derivatives so this would be a circular argument. – 2012-09-20
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0@crf yea, but as DonAntonio said, it would be a circular argument – 2012-09-20
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0Why would you want to avoid using chain rule? – 2012-09-20
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0The chain rule is your friend... – 2012-09-20
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0@AlexBecker because the problem becomes trivial with chain rule, and I like thinking :) – 2012-09-20
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0@AlexBecker if there is no way to do it without chain rule, then thats fine. I was just wondering. Its been on my head for a while now. – 2012-09-20
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25What is your definition of $e^x$? – 2012-09-20
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3In my Analysis class, we defined $e^x$ as the solution of $f'(x) = f(x)$ with $f(0) = 1$. So um, that works. – 2012-09-20
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7How do you do this *with* the chain rule? – 2012-09-20
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0This question has been asked recently, but I can't find it... – 2012-09-20
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4@ChrisEagle let $y=e^x$ then $\ln(y)=x$ hence $\frac{1}{y}y'=1$ thus $y'=y$ aka $\frac{d}{dx}e^x=e^x$ – 2012-09-20
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1@James: So defining $e^x$ as the inverse of the integral of $1/x$? How perverse. – 2012-09-20
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0$e^x=\cosh x + \sinh x, \cosh'=\sinh,\sinh'=\cosh$ – 2012-09-20
5 Answers
When using the definition $$\mathrm e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n$$ you can proceed as follows: $$\begin{aligned} \frac{\mathrm d}{\mathrm dx}\mathrm e^x &= \lim_{h\to 0}\frac{\mathrm e^{x+h}-\mathrm e^x}{h}\\ &= \lim_{h\to 0}\frac{\lim\limits_{n\to\infty}\left(1+\frac{x+h}n\right)^n - \lim\limits_{n\to\infty}\left(1+\frac xn\right)^n}{h}\\ &= \lim_{h\to 0}\lim_{n\to\infty}\frac{\left(1+\frac{x+h}n\right)^n - \left(1+\frac xn\right)^n}{h} \end{aligned}$$ Now $$\left(1+\frac{x+h}{n}\right)^n = \sum_{k=0}^n{n\choose k}\left(\frac{h}{n}\right)^k\left(1+\frac{x}{n}\right)^{n-k}$$ and therefore $$\begin{aligned} \frac{\mathrm d}{\mathrm dx}\mathrm e^x &= \lim_{h\to 0}\lim_{n\to\infty}\sum_{k=1}^n{n\choose k}\frac{h^{k-1}}{n^k}\left(1+\frac{x}{n}\right)^{n-k}\\ &= \lim_{h\to 0}\left(\lim_{n\to\infty}{n\choose 1}\frac{1}{n}\left(1+\frac{x}{n}\right)^{n-1}+h\lim_{n\to\infty}\sum_{k=2}^n{n\choose k}\frac{h^{k-2}}{n^k}\left(1+\frac{x}{n}\right)^{n-k}\right)\\ &= \lim_{n\to\infty}{n\choose 1}\frac{1}{n}\left(1+\frac{x}{n}\right)^{n-1}+\lim_{h\to 0}h\lim_{n\to\infty}\sum_{k=2}^n{n\choose k}\frac{h^{k-2}}{n^k}\left(1+\frac{x}{n}\right)^{n-k}\\ &= \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^{n-1}\\ &= \lim_{n\to\infty}\frac{\left(1+\frac{x}{n}\right)^{n}}{1+\frac{x}{n}}\\ \end{aligned}$$ Since the limit for numerator and denominator exists independently, we can calculate them separately. The numerator is just the definition of $\mathrm e^x$, and the limit of the denominator is $1$, so we arrive at $$\frac{\mathrm d}{\mathrm dx}\mathrm e^x = \mathrm e^x$$
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0Not following the first line after "and therefore" (in the middle). According to the line before it, shouldn't we have the fraction $\cfrac{h^{k-1}}{n^k}$, rather than $\cfrac{h^{k-1}}k$? Also, observe that you passed the $\lim\limits_{n\to\infty}$ right through the $\sum_{k=1}^n$ on the line after that. That's not good.... – 2012-09-20
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0@CameronBuie: Oops, you're right. Also, I now notice that I've forgotten the binomial coefficients in the sum. I'll try to fix it. – 2012-09-20
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0@CameronBuie: Now better? – 2012-09-20
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1From line 3 to line 4 in the last part you not really valid; you'd need to show that the inner limit exists, or at least that the expression in it stays bounded... – 2013-05-03
Define $e$ implicitly by $\lim_{h \rightarrow 0} \frac{e^h-1}{h}=1$. Calculate, $$ \frac{d}{dx} e^x = \lim_{h \rightarrow 0} \frac{e^{x+h}-e^x}{h} = e^{x}\lim_{h \rightarrow 0} \frac{e^h-1}{h} = e^x.$$ This definition assumes that properties of exponential functions are somehow known.
In contrast, the definition that defines the $\ln(x) = \int_{1}^{x} \frac{dt}{t}$ allows you to derive properties of the natural log. Then the exponential function is introduced as the inverse function and its properties can be induced from those already proven for the natural log.
Logically the definition of the natural log as primary has advantages. But, pedagogically if you wish to discuss the exponential function before integral calculus then some sort of chicanery is required.
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1But isn't the hardest part left? Namely demonstrating that $\lim_{h\to 0}\frac{e^h-1}{h}=1$... – 2015-09-20
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2@BartPatzer well, if you study $y=2^x$ and $y=3^x$ it is clear these have tangent lines at $x=0$ and it is clear that the slopes are $0.69..$ and $1.097..$ hence as there is a continuous selection of bases for $y=a^x$ between $a=2$ and $a=3$ it is reasonable to suppose such a base $a$ exists as to produce slope $1$ at $x=0$. The implicit definition of $e$ I state here is merely the formulation of the slope of the tangent for such a base. .. – 2015-09-20
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0The demonstration that every value between $\ln(2)=0.69..$ and $\ln(3) =1.097...$ is attained as the slope of one and only one base $a$ is a delicate question beyond the scope of calculus 1. But, this is no worse than the supposition that the formula $f(x)=a^x$ is well-posed for $x$ a real value. So, it's a reasonable definition. Of course, the power series definition for $e$ is more logically economical. – 2015-09-20
Hmmm.... Well, how precisely have you defined $e^x$? Depending on the answer, the approach will vary. If you've defined $$e^x=\sum_{k=0}^\infty\frac{x^k}{k!},$$ then it will follow fairly readily that $e^x$ is its own derivative, using Taylor series properties.
If on the other hand you've defined $$e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n,$$ then you may have a slightly harder way to go. I think using the difference of $n$th powers formula may help.
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0I think you mean to use an exponent of $n$, not $1/n$, in the second definition. – 2012-09-20
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0You don't need to interchange limits, see my answer. – 2012-09-20
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0I've seen it, celtschk, and I'm not altogether convinced, yet. Let me know when you've addressed the issues I pointed out. – 2012-09-20
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0Note that my corrected version still needs no interchange of limits. – 2012-09-20
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0I did note that. Thanks! – 2012-09-20
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1How do you know that you can take the derivative of that series by taking the derivatives of its terms? That generally *does* require special properties of power series, I believe. – 2013-07-15
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0@dfeuer: Oops! You're right. I'm not sure why I thought I could get away with limit interchange without any previous results suggesting it was possible. Thanks! – 2013-07-15
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0@downvoter: Care to comment on what you feel my answer is lacking, so that I can rectify it? – 2017-08-21
To follow up on "what is your definition of $e^x$", if your definition is as the solution to the differential equation $y'=y$ such that $y(0)=1$, then you have nothing to prove!
If the definition of $e^x$ is "the differentiable solution to $f(x+y) = f(x)f(y)$ with $f'(0) =1$, this way works:
Putting $y = 0$, $f(x) = f(x)f(0)$ for all $x$, so $f(0) = 1$.
$(f(x+h)-f(x))/h = (f(x)f(h)-f(x))/h = f(x)(f(h)-1)/h = f(x)(f(h)-f(0))/h $. Taking the limit as $h \to 0$, $f'(x) = f'(0)f(x)$.
We now can use the differential equation approach.
Note: If this seems familiar, I have used this answer previously in a similar context.
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0how do you work backwards from that to show that $f^{\prime}(0)=1$? I'm currently trying to figure that out here: http://math.stackexchange.com/questions/1668026/proof-of-existence-and-uniqueness-of-the-exponential-function-using-odes – 2016-02-23
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0Because $f'(0) = \lim_{h \to 0} (f(h)-f(0))/h$. – 2016-02-23
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0but if I am trying to prove that $f^{\prime}(0) = 1$, how do I know that $\lim_{h \to 0} (f(h)-f(0))/h = 1$?? I can't use the thing I'm trying to prove in order to prove it. – 2016-02-23