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I am working on proving different mathematical formulas. I am currently working on proving that $\frac{d}{dx} e^x = e^x$ . This is my proof so far:

$$\frac{d}{dx} e^x = \lim_{h \to0}\frac{e^{x+h} - e^{x}}{h} = $$ $$ \lim_{h \to 0} \frac{e^x(e^h-1)}{h} = e^x \cdot\lim_{h \to 0} \frac{(e^h-1)}{h}$$ Need to prove that $\lim_{h \to 0} \frac{(e^h-1)}{h} = 1$.

Using substitution: $h = \ln(t+1)$

$$\lim_{t \to 0}\frac{t}{\ln(t+1)}$$

$$\lim_{t \to 0}\frac{t \cdot \frac{1}{t}}{\ln(t+1)\cdot \frac{1}{t}}$$

$$\lim_{t \to 0}\frac{1}{\ln(t+1)^{\frac{1}{t}}}$$

$$\frac{1}{\ln(e)} = \frac{1}{1} = 1$$

$\blacksquare$

Am I allowed to say $t \to 0$ in a proof when it is actually $\ln(t+1) \to 0$? Do you have any corrections on the proof?

Michael Hardy
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peanut_butter
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    There is a question like yours: http://math.stackexchange.com/questions/190773/proof-of-fracddxex-ex?rq=1 – Leafar Feb 06 '15 at 17:59
  • You are applying this property of limits. – Pp.. Feb 06 '15 at 18:01
  • $$\lim_{h \to 0} (e^h-1)/h = \lim_{h \to 0} ([(1+h)^{1/h}]^h - 1)/h = \lim_{h \to 0} (1+h - 1)/h = 1$$ – Balarka Sen Feb 06 '15 at 18:50
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    No problem for the last line but my question is that how do you know $lim_{n\to \infty}(1+\frac1n)^n=e$ is more evident than $(e^x)'=e^x$ ?? – Arashium Feb 06 '15 at 18:53
  • $e = \lim_{n \to 0} (1 + n)^{1/n}$ is the definition of $e$. – Balarka Sen Feb 06 '15 at 18:54
  • @Arashium It is usually easy to prove that $(1+\frac{1}{n})^n$ is convergent: Using the Bernoulli inequality you can prove that the sequence is monotonic and bounded.... To define $(e^x)'=e^x$ you need to prove that the exponential functions $a^x$ are differentiable (which is "intuitively" clear, but hard to prove, and the intuitively clear comes from the fact that we draw it in a way which shows that it is differentiable, so circular logic) AND that the derivative is continuous as function of $a$... – N. S. Feb 06 '15 at 19:10
  • @BalarkaSen as far as know, e is defined as $exp(1)$ or the number in which $\ln(x)=1$. It depends on how you define exp and logarithm. But if you define e as such limit you make everything very complicated in other aspects. – Arashium Feb 06 '15 at 19:14
  • @N.S. Could you tell me how you define $a^x$? Do you define it before $\exp(x)$? – Arashium Feb 06 '15 at 19:15
  • @Arashium And how do you propose to define $\ln$? The limit definition of $e$ is standard, as far as I know. What you're trying to do sounds suspiciously tautological. – Balarka Sen Feb 06 '15 at 19:22
  • @Arashium: $a^q$ for rational $q$ is defined by the only way it can be in order to keep the rules for exponents consistent. Then $a^x$ is the continuous extension of $a^q$. – Matthew Leingang Feb 06 '15 at 19:23
  • @Arashium You define $a^r$ for all rational numbers and you prove that it extends to a continuous function...While the approach with defining first $\ln (x)$ and using the IVT to define $e$ makes things easier indeed. But the reason why it makes things easier is because it uses strong results from Analysis, you need to prove a lot of things before defining $e$....And of course the more you know the simpler things become. – N. S. Feb 06 '15 at 19:25
  • @N.S. this definition solves the problem for real number but it cannot be extended to complex numbers. – Arashium Feb 06 '15 at 19:33
  • @Arasthehium To compare: the limit definition of $e$ only requires the definition of convergent sequences and the knowledge that every bounded monotonic sequence is convergent, which is an immediate lemma if $\mathbb R$ is properly defined. To define $e$ the "easy" way, you need to speak first about continuity, you need to prove the IVT (which uses implicitly in the proof the convergence of monotonic bounded sequences), to introduce the Riemann sums and Riemann integrals and to prove that every continuous function is Riemann integrable. Moreover, you need to prove that the integral from $1$ .. – N. S. Feb 06 '15 at 19:34
  • ...to $x$ is continuous, which is easy but still needs to be proven. – N. S. Feb 06 '15 at 19:34
  • @N.S. I dont think there is any function more innocent than $\frac1x$ for Riemann sums :). But you referred to a beautiful point that these definitions are not so far from each other. – Arashium Feb 06 '15 at 19:38
  • @Arashium What do you mean by that? To define the complex exponential all you need is to know what $e^x$ is as a real exponential. – N. S. Feb 06 '15 at 19:39
  • @Arashium Even with the innocence of $\frac{1}{x}$, I would still prefer to define $e$ and the exponential the "non-intuitive" way at the beginning of an introductory Analysis class, than wait until the end to be able to define it.... Again the point I am making is that many things become "easier and more intuitive" after you cover much more material. Trigonometric functions are much easier to understand once one knows the complex exponential, should we introduce them to secondary school students this way? – N. S. Feb 06 '15 at 19:45
  • @N.S. I absolutely agree with you about the level of students. – Arashium Feb 06 '15 at 19:49

4 Answers4

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For exponential you cannot use Taylor series or L'Hopital since you will stuck into recursive reasoning.

Note that exponential is introduced in math after natural logarithm. So use function inverse:

$$y=\exp(x)$$

$$x=\ln(y)$$

$$\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}=\frac{1}{\frac{1}{y}}=y=\exp(x)$$

Additional explanation

Definition of $\ln (x)$:

$$\ln(x)=\int_1^x\left(\frac1t\right)dt$$

And $\exp(x)$ is defined as inverse function of $\ln(x)$.

Arashium
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That $e$ is the unique number $a$ for which $\displaystyle\lim_{h\to0}\frac{a^h-1}h$ is equal to $1$ is sometimes taken to be the definition of $e$.

Notice that with $a=1$ we can see that the limit is less than $1$, as follows: As $h$ goes from $0$ to $1$, $a^h=2^h$ goes from $1$ to $2$, so the slope of that secant line is $1$. Since the tangent line gets steeper as you go from left to right, the slope of the tangent line at $0$ must therefore be less than $1$. You can also show it's more than $1/2$ by considering the secant line through $h=0$ and $h=-1$.

Using $h=0$ and $h=-1/2$, you can show in the same way that $\displaystyle\lim_{h\to0}\frac{4^h-1}h$ is bigger than $1$.

Thus the number $a$ for which $\displaystyle\lim_{h\to0}\frac{a^h-1}h$ is exactly $1$ must be somewhere between $2$ and $4$.

There are many characterizations of $e$. Which of them is taken to be the definition depends on context and is a judgment call.

Michael Hardy
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    Personally I really don't like this definition at all. It assumes that all the exponential functions are differentiable at $0$, and that the derivative of $a^x$ at $0$ is continuous as function of $a$ (therefore the continuity of $\ln(x)$(... The funny part is that after defining $e$ this way, many textbooks go on to "prove" that the exponentials are differentiable at $0$ and that $\ln$ is continuous.... – N. S. Feb 06 '15 at 19:55
  • @N.S. : Vast numbers of students majoring in poetry, music, pre-med, sociology, business administration, jurisprudence, etc., who will never be prepared to appreciate logical rigor, are pressured into taking calculus. It is worse than a waste of time to expect them to appreciate rigor. Such courses degenerate into teaching students that mathematics is just a bunch of algorithms, plus incomprehensible stuff they have to sit through to earn a grade. Rigor has its place but that audience isn't it. ${}\qquad{}$ – Michael Hardy Feb 07 '15 at 00:12
  • I know, but this approach is basically saying in a convoluted way that $(e^x)'=e^x$. For those classes, we should simply define $e$ this way... – N. S. Feb 07 '15 at 00:31
  • @N.S. It is saying that the derivative of an exponential function is a scalar multiple of the exponential function, and then that there's only one base for which that scalar is $1$. ${}\qquad{}$ – Michael Hardy Feb 07 '15 at 00:37
  • . . . and beyond saying that that rigor is worse than wasted on that audience, I should add that they have never suspected that math is not a dogmatic subject in which one should just believe and obey, nor that there is any satisfaction in coming to the point where one understands something instead of having to take the word of the instructor and the textbook, and they resent being treated as if they do understand that. Moreover, one squanders an opportunity to tell them that calculus is a great intellectual, scientific, aesthetic and cultural achievement that$\ldots\ldots$ – Michael Hardy Feb 07 '15 at 00:40
  • . . . . . has been and continues to be influential in the course of history. – Michael Hardy Feb 07 '15 at 00:41
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A meta-answer: Each of the three answers so far are coming from a different definition of the exponential function. They are all equivalent. But in each course, only one of them is the definition.

So it's important to understand what the definitions are in your course before you can prove this statement yourself.

Matthew Leingang
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$$\frac{d}{dx}e^x=\frac{d}{dx}\sum_{k=0}^\infty \frac{x^k}{k!}\underset{(why\ ?)}{=}\sum_{k=0}^\infty \frac{d}{dx}\left(\frac{x^k}{k!}\right)=\sum_{k=1}^\infty \frac{x^{k-1}}{(k-1)!}\underset{\ell=k-1}{=}\sum_{\ell=0}^\infty \frac{x^\ell}{\ell !}=e^x.$$

idm
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