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I have a problem : How to prove $$\dfrac{d}{dx}e^x=e^x\;?$$

My answer is \begin{eqnarray} \dfrac{d}{dx}e^x&=&\lim\limits_{h\to 0}\dfrac{e^{x+h}-e^x}{h}\\ &=&\lim\limits_{h\to 0}\dfrac{e^{x}e^h-e^x}{h}\\ &=&e^{x}\lim\limits_{h\to 0}\dfrac{e^h-1}{h}\\ \end{eqnarray} But I don't know to find $\lim\limits_{h\to 0}\dfrac{e^h-1}{h}$. So how to find it?

Note: I cannot use L'Hospital rule.

Jam
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Ongky Denny Wijaya
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    What definition of e^x are you using here? – SmileyCraft Dec 16 '18 at 12:10
  • @ SmileyCraft e^x = exponential – Ongky Denny Wijaya Dec 16 '18 at 12:25
  • I don't think that Smiley needed to know that but there are multiple possible definitions of the function and no official one. One of those possible definitions is that it is the solution of that differential equation and, with that, there would be nothing to prove. A popular one is the power series which would also make this easy. – badjohn Dec 16 '18 at 12:28
  • It depends on your definition of $e$ (or $e^x$). If you have $e = \lim_\limits{h \to \infty}\big(1+\frac{1}{h}\big)^h = \lim_\limits{h \to 0}(1+h)^{\frac{1}{h}}$, then $\lim_\limits{h \to 0} e^h = 1+h$, and the limit becomes $1$. – KM101 Dec 16 '18 at 12:31
  • @ Jam, using the power series of $e^x$ – Ongky Denny Wijaya Dec 16 '18 at 12:33
  • @OngkyDennyWijaya Then you can use this question. Try to Google questions like this if you can since most of them have been asked before and it'll save you (and us) time. Also, your English is fine, there's no need to put yourself down :) If we don't understand anything, I'm sure we can let you know. – Jam Dec 16 '18 at 12:35

3 Answers3

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The key fact is that

$$\lim_{n\to \infty} \left(1+\frac1n\right)^n \to e \implies \lim_{n\to \infty} \left(1+\frac1x\right)^x \to e$$

indeed for any $x\in (n,n+1)$

$$\left(1+\frac1{n+1}\right)^n \le \left(1+\frac1x\right)^x \le \left(1+\frac1{n}\right)^{n+1} $$

then take the limit and use squeeze theorem.

Then

$$\lim_{x\to \infty} \left(1+\frac1x\right)^x \to e \implies \frac{\log \left(1+\frac1x\right)}{\frac1x}\to 1 \implies \frac{\log (1+t)}{t}\to 1 \, t\to 0$$

then by $h=\log (1+t)\to 0, t \to 0$

$$\lim\limits_{h\to 0}\dfrac{e^h-1}{h}=\lim\limits_{t\to 0}\dfrac{t}{\log (1+t)}=1$$

user
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  • How do you go from the limit of a sequence to the limit of a function? $\lim_{n\to\infty}\sin(\pi n)=0$ (sequential limit), but definitely $\lim_{x\to\infty}\sin(\pi x)$ ($x\in(0,\infty)$) is not $0$, as it doesn't even exist. – egreg Dec 16 '18 at 12:19
  • @egreg Are you referring to $\frac{\log \left(1+\frac1n\right)}{\frac1n}\to 1 \implies \log (1+t)/t =1 , t\to 0$? – user Dec 16 '18 at 12:25
  • @egreg I've added all the details for the derivation form the foundamental limit for e. – user Dec 16 '18 at 12:34
  • Thanks for your answer.. – Ongky Denny Wijaya Dec 16 '18 at 12:36
  • @OngkyDennyWijaya You are welcome. The answer depends upon the definition for $e^x$ you are allowed to use. From the definition by series the proof is straightforward. The way I've indicated is the one from the definition of $e$ as a limit of a sequence (sometimes that definition comes first that one by series). – user Dec 16 '18 at 12:43
  • @gimusi How do you prove that $x\mapsto(1+x^{-1})^x$ is increasing? FYI, I'm just playing the devil's advocate. – egreg Dec 16 '18 at 13:54
  • @egreg I appreciate that really and you are right, I fix that step :) – user Dec 16 '18 at 14:33
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Personally, the simplest and best way to prove this identity is not from the definition of derivative, but to use the power series for $e^x$:

$$e^x = \sum_{k=0}^\infty \frac{x^k}{k!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{2 \cdot 3} + ...$$

Since

$$\frac{d}{dx} \left( f(x) + g(x) \right) = \frac{d}{dx} f(x) + \frac{d}{dx} g(x)$$

performing the derivative becomes quite easy and should easily show the identity.

The power series will also prove useful if going from the definition of derivative as a limit as well, if you insist on using that. Personally, the method above I find much more elegant and intuitive but that's a matter of opinion.

Note that interchanging the operations of derivatives and infinite summation does not always hold, i.e.

$$\frac{d}{dx} \sum_{k=0}^\infty f(k) = \sum_{k=0}^\infty \frac{d}{dx} f(k)$$

need not always hold. (While the derivative operation does "distribute" per the summation rule above, that is for finite summation, and need not hold true when we have infinite summation. That it does is a consequence of other facts.)

My own personal understanding of the underlying reason is lacking, so I'm mostly deferring to Riley's comments on my answer in the matter. For this to be applicable in general, the sequence of derivatives for $f(k)$ must converge uniformly - which, in this case, happens to be true for the polynomials $f(k) = x^k/k!$.

Edit

The proof that a power series with uniform convergence may be differentiated termwise within its radius of convergence is given as theorem 2 on this page and on this page. The proof that $e^x$ has a radius convergence of $\infty$ (and hence is termwise differentiable everywhere) is given on this page.

Community
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PrincessEev
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    The only issue I have with this is that although it is probably more enlightening, it requires that differentiation and infinite summation can be interchanged. – Riley Dec 16 '18 at 12:18
  • I'm confused. Can they not be interchanged owing to how the derivative effectively distributes across sums? (Not sure if distributes is the right word but you probably get what I mean.) – PrincessEev Dec 16 '18 at 12:20
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    Differentiation commutes with finite summation, not necessarily with infinite summation. – Riley Dec 16 '18 at 12:23
  • Thanks for your answer.. – Ongky Denny Wijaya Dec 16 '18 at 12:36
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    I think I should reiterate this answer is incomplete as it stands. Interchanging differentiation and infinite summation is permitted in this case because a sufficient condition is the uniform convergence of the derivatives, and the exponential does converge uniformly. Without these two facts, this answer suggests that a statement true for finite sums is true for infinite sums which is false. – Riley Dec 16 '18 at 13:11
  • There's only so much I can do when I have little understanding of the topic I'm being told to complete the answer with. I edited in some details but it is mostly parroting what you said. If someone more knowledgeable on the matter wants to expound upon it in more detail, they're welcome to edit my post further. – PrincessEev Dec 16 '18 at 13:25
  • @EeveeTrainer I've edited your answer to include links to the proofs of the relevant theorems. I hope that's okay. – Jam Dec 16 '18 at 19:51
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\begin{eqnarray} e^x&=&\sum\limits_{k=0}^\infty \dfrac{x^k}{k!}\\ \dfrac{d}{dx}e^x&=&\dfrac{d}{dx}\sum\limits_{k=0}^\infty \dfrac{x^k}{k!}\\ &=& \sum\limits_{k=0}^\infty \dfrac{d}{dx}\dfrac{x^k}{k!}\\ &=& \sum\limits_{k=1}^\infty \dfrac{kx^{k-1}}{k!}\\ &=& \sum\limits_{k=1}^\infty \dfrac{x^{k-1}}{(k-1)!}\\ &=& \sum\limits_{k=0}^\infty \dfrac{x^{k}}{k!}\\ &=&e^x \end{eqnarray}

Ongky Denny Wijaya
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