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Let $f(x) = e^x$

The derivative is by definition: $$\frac{d}{dx}e^x = \lim_{h\rightarrow 0}\frac{e^{x+h} -e^{x}}{h} =e^x\lim_{h\rightarrow 0}\frac{e^h-1}{h}$$

Using L'Hospital rule: $$\lim_{h\rightarrow 0}\frac{e^h-1}{h}=1$$

Therefore, $$\frac{d}{dx}e^x = e^x$$

My problem with this proof is that the L'Hospital rule already implies knowing how to calculate the derivative of the exponential. Thus, this kind of proof looks to me like a circular reasoning.

Is there a proof that shows how $\frac{d}{dx}e^x = e^x$ without having to calculate this derivative in any of the intermediate steps?

AWanderingMind
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2 Answers2

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You're correct, it is circular reasoning. All of this depends on how you define the function $e^x$. The most popular in higher mathematics, is the power series definition:

\begin{equation} \exp(z) = \sum_{k = 0}^\infty \frac{z^k}{k!} \end{equation}

This function is defined for all complex numbers. If you know a few facts about uniform convergence, then you'll know that the derivative of the above summation is obtained by differentiating each term of the summation(w.r.t z). You will see you'll again get the same power series if you differentiate.

General Grievous
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  • This is also the historical approach (except it was for defined as a real power series) and the most convincing (most "proofs" use $\ln$ so this is just reporting the pb on $\ln$ derivative, which somehow is circular argument too). – zwim Jun 20 '20 at 10:30
  • But there are other ways. You can define it as the unique solution to the differential equation OP mentioned. But this is the most flexible one. – General Grievous Jun 20 '20 at 10:38
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By the conventional, high-school definition $$ e^{\pm h} = \lim_{n\to +\infty}\left(1\pm\frac{h}{n}\right)^n $$ and $e^h\geq 1+h$ holds true for any $h\in\mathbb{R}$. This implies $e^{-h}\leq \frac{1}{1+h}$ for any $h>-1$ and $e^h\leq\frac{1}{1-h}$ for any $h<1$. It follows that the limit $\lim_{h\to 0}\frac{e^h-1}{h}$, leading to the derivative of $e^x$, can be simply computed by squeezing, since

$$ \lim_{h\to 0}\frac{(h+1)-1}{h} = 1 = \lim_{h\to 0}\frac{\frac{1}{1-h}-1}{h}.$$

Jack D'Aurizio
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