Two Proofs Of The Unique Self-Differentiation Property of CexCe^x

Ever since I learned elementary calculus, I have been wondering about the proof of the unique self-differentiation of CexCe^x, CC a constant. One of the proofs below I was led to by a calculus textbook; another I figured out by myself.

11: Using the Mean Value Theorem

Let f(x)f(x) be a function so that f(x)=f(x)f(x)=f'(x). Define a new function

g(x)=f(x)ex(1)g(x)=\frac{f(x)}{e^x} (1)

Now, find g(x)g'(x):



But our base assumption is that f(x)=f(x)f(x)=f'(x), or that f(x)f(x)=0f'(x)-f(x)=0. In other words,


Now, the Mean Value Theorem implies that only constant functions have zero derivative. This means that g(x)=Cg(x)=C for some constant CC. Plugging this into (1)(1) yields:



22: Using the Taylor Expansion of exe^x

Note: I would like to thank Alexander Gibson for pointing out that this proof only applies to the set of analytic functions.

Once again, let f(x)f(x) be a function so that f(x)=f(x)f(x)=f'(x). Notice that this implies the following:



f(x)=f(x)    limh0f(x+h)f(x)h=limh0f(x+h)f(x)h    f(x)=f(x)f(x)=f'(x)\implies\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}=\lim_{h\to 0} \frac{f'(x+h)-f'(x)}{h}\implies f'(x)=f''(x)

and so forth. We can also say that


for some constant CC. Use the theory of Taylor series to expand f(x)f(x):


for f(x)f(x) around any point x=ax=a. Inputting our initial assumption into the Taylor expansion of f(x)f(x) yields:


Divide both sides by CC:


Notice that the right hand side is the Taylor expansion of exe^x about a point x=ax=a. We can conclude that



Note: It was okay to assume that the Taylor series converges, because it always does in a small interval about xax-a.

Whew! That was a lot of theory. If you want to share your own proof of the unique self-differentiation property of CexCe^x, then share it down below in the comments.

Note by Andrei Li
1 year, 11 months ago

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Here is a derivation for Euler's number, based on the self-differentiation property. This is a less ambitious derivation than the one posted in the note. It doesn't attempt to prove the uniqueness of the self-differentiation property.

Suppose the following is true for some constant AA over all xx.

ddxAx=Ax\large{\frac{d}{dx} A^x = A^x}

Evaluate the difference quotient:

ddxAx=Ax+dxAxdx=AxAdx1=dxA=(1+dx)1/dx\large{\frac{d}{dx} A^x = \frac{A^{x + dx} - A^x}{dx} = A^x \\ A^{dx} - 1 = dx \\ A = (1 + dx)^{1/dx}}

The limit of this expression as dxdx approaches zero is Euler's number.

Steven Chase - 1 year, 11 months ago

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It's a clever proof! Not what I was expecting, but it's smart!

Andrei Li - 1 year, 10 months ago

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I was attempting something less ambitious (simply a derivation of Euler's number). I have made notes explaining that.

Steven Chase - 1 year, 10 months ago

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@Steven Chase Okay! No problem. ;)

Andrei Li - 1 year, 10 months ago

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y=dydxy = \frac{dy}{dx}

1ydy=dx\frac{1}{y}dy = dx

1ydy=dx\int \frac{1}{y}dy = \int dx

lny=x+C1\ln y = x + C_1

elny=ex+C1e^{\ln y} = e^{x + C_1}

y=exeC1y = e^{x}e^{C_1}

y=Cexy = Ce^x

David Vreken - 1 year, 10 months ago

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Toke dey ke? lol.

Arkajyoti Banerjee - 1 year, 10 months ago

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The second proof doesn't necessarily follow, you've assumed the taylor series of f(x) converges, or in other words, that f(x) is what's called analytic.Some functions can be infinitely differentiable(called smooth) and still not be analytic, for instance e^(-1/x), which has all its derivatives at 0 equal to 0, and yet it eventually grows to be much bigger than 0.

However, luckily, in complex analysis, this distinction completely dissolves, and it turns out that even if just one derivative exists, the entire function is analytic, which is just one example of why everything is better in the complex case.

Alexander Gibson - 1 year, 10 months ago

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Thank you for your advice! I have generalized my proof for any interval about aa, where aa is any point along the xx-axis.

Andrei Li - 1 year, 10 months ago

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Nope, not even that works, some dastardly functions are not analytic no matter where they are centered around

Alexander Gibson - 1 year, 10 months ago

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@Alexander Gibson (

Alexander Gibson - 1 year, 10 months ago

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@Alexander Gibson Thank you for your advice! I have now included a note that this proof only works for analytic functions.

Andrei Li - 1 year, 10 months ago

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Isn't the first proof some sort of a circular argument?

Rahul Sethi - 1 year, 1 month ago

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