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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.

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    $e^x=\cosh x + \sinh x, \cosh'=\sinh,\sinh'=\cosh$2012-09-20

5 Answers 5

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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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    From 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
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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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    The 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
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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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    @downvoter: Care to comment on what you feel my answer is lacking, so that I can rectify it?2017-08-21
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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!

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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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    but 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