I have been trying to differentiate the exponential function from first principles without the use of Taylor's series or the derivative of its inverse function ($\frac{d}{dx} (\ln x) = \frac{1}{x}$ and $\ln (e^x) = x$.
Let $f(x) = e^x$, then differentiating $f(x)$ from first principles,
$$f^\prime(x) = \lim_{\delta x \to 0} \frac{f(x+\delta x) – f(x)}{\delta x} = \lim_{\delta x \to 0} \frac{e^{x+\delta x} – e^x}{\delta x} = \lim_{\delta x \to 0}\frac{e^x(e^{\delta x} – 1)}{\delta x} = e^x \lim_{\delta x \to 0} \frac{e^{\delta x} – 1}{\delta x}$$
Therefore, in order to prove from first principles that $\frac{d}{dx}(e^x) = e^x$, I would need to first show that
$$\lim_{\delta x \to 0} \frac{e^{\delta x} – 1}{\delta x} = 1$$
However, I am not sure how to evaluate this limit and the use of L'Hôpital's rule requires preliminary knowledge on the derivative of $e^x$.
Is it possible to prove the derivative of $e^x$ from first principles solely using limits, or is it impossible as the knowledge of its derivative is a prerequisite to its discovery by Bernoulli?
Best Answer
As I've said fairly often in the last few days (for some reason), one of my favorite equations is: $$e^x\ge x+1$$ The reason, partly, is that it uniquely defines $e$ without calculus. Hint for a proof: use this. (By the way, do equations need equals signs? Or is it equalities that need equals signs?)
Now, replacing $x$ by $-x$, we get $e^{-x}\ge1-x$, so: $$e^x\le\frac1{1-x}$$ (The inequality gets reverse for $x>1$, as the right-hand side is negative there. But we only care about when $x$ is near zero.)
Thus: \begin{align} x+1\le{}&e^x\le\frac1{1-x}\\ x\le{}&e^x-1\le\frac x{1-x}\\ 1\le^*{}&\frac{e^x-1}x\le^*\frac1{1-x} \end{align} *Since we just divided by $x$, the inequalities get reversed if $x$ is negative. It doesn't affect the argument.
Let $x$ tend to zero. By the squeeze theorem: $$1=\lim_{x\to0}\frac{e^x-1}x$$