$\frac{db^x}{dx}$ without $e$

Question

For no other reason other than interest, I'm trying to find the general derivative of $b^x$ without using a definition of $e$ from a different context.

I feel like, chronologically in history, this would have been the first time $e$ would have popped up in the context of calculus.

Every proof of $\frac{db^x}{dx}$ I can find uses the result of $\frac{de^x}{dx}=e^x$. But at the time (and correct me if I'm wrong), $e$ wasn't really popularized. It was (almost) being used in disguise by Napier, only because $(1-10^{-7})^{10^7} \approx e^{-1}$. When Netwon came around, Bernoulli may have been looking to find the value of $\lim_{n\to \infty}(1+1/n)^n$, but I don't see any motivation to consider

$$\frac{d\left(\lim_{n\to \infty}(1+1/n)^n\right)^x}{dx}$$

before the general case $\frac{db^x}{dx}$. I'm sure along the way of finding the derivative of $b^x$, a clear motivation for defining $e$ will pop-up .. but I'd like to find a proof that starts off assuming no prior knowledge of $e$.

If you start from definition, you very quickly arrive at

$$\frac{db^x}{dx} = b^x \lim_{h\to 0}{\frac{b^h-1}{h}}$$

but from here I'm stuck. How to show $\exists c \in \Re$ such that

$$\lim_{x \to 0}\frac{c^x-1}{x}=1$$

to proceed?

Answer 1

\begin{align} \frac d {dx} b^x = \lim_{\Delta x\to0} \frac{\Delta b^x}{\Delta x} = \lim_{\Delta x\to 0} \frac{b^{x+\Delta x} - b^x}{\Delta x} = {} & \underbrace{ \lim_{\Delta x\to0} \left( b^x \frac{b^{\Delta x} - 1}{\Delta x} \right) = b^x \lim_{\Delta x\to0} \frac{b^{\Delta x} - 1}{\Delta x} }_{\begin{array}{l} \large\text{This works because $b^x$ is “constant'' in} \\ \large\text{the sense that it doesn't change as $\Delta x$} \\ \large\text{approaches $0.$} \end{array}} \\[15pt] = {} & \big( b^x \cdot \text{constant} \big) \qquad \text{But this “constant''} \\[2pt] & \text{is “constant'' in the sense that is doesn't} \\ & \text{change as $x$ changes.} \end{align}

Now suppose we can show that the "constant" at the end is less than $1$ if $b=2$ and greater than $1$ if $b=4.$ One would then expect that for some number $b$ between $2$ and $4,$ the constant is $1,$ and that number between $2$ and $4$ is $e.$

For $b=2,$ consider the points $(0,b^0) = (0,1)$ and $(1,b^1) = (1,2).$ The slope of the secant line is $1$ and should be more than the slope of the tangent line at $x=0$ since the curve gets steeper as you go to the right.

For $b=4,$ do likewise with $(-1/2,b^{-1/2}) = (-1/2,1/2)$ and $(0,b^0) = (0,1),$ and conclude that $4$ is too big to be $e.$

Finally, writing $b^x= e^{x\log_e b}$ and applying the chain rule, one concludes that the "constant" is $\log_e b.$

Answer 2

I wish to suggest one.

Recall the Napierian logarithm definition: For all real $y > 0$ we define $$\log y = \lim_{n \to \infty +}n(y^{1/n} - 1)$$.

Let $c > 0$ be a real number. Then $$\log c = \lim_{n \to \infty +}n(c^{1/n} - 1) = \lim_{x \to 0}\frac{1}{x}(c^{x} - 1) = \frac{d}{dx}c^{x} \mid_{x=0}.$$ Choose $c = e$, qed.

Answer 3

Using real analysis - which did not exist at the time of Napier - we could talk about the existence of the constant in your $h \to 0$ limit:

$$ \frac{db^x}{dx} = b^x \lim_{h\to 0}{\frac{b^h-1}{h}}$$

Notice that $h$ is a very small number and your equation suggests for some constant $c$ :

$$ b^h \approx 1 + hc$$

What happens when we multiply two of these factors, using what we know about the exponential function?

$$ b^h \cdot b^{h'} \approx (1 + hc)(1+ h'c) = 1 + (h+h')c+o(h^2) \approx b^{h + h'}$$

This is how we get that very small compounded interest rates can be added.

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Hmm... we could do it the other way around

$$ a^h \cdot b^h \approx (1 + h c_1)(1 + h c_2) \approx 1 + h(c_1 +c_2) + o (h^2) \approx (ab)^h $$

Eventually $c_1 = \log a$ and $c_2 = \log b$ and we are showing that $c_1 + c_2 = \log (ab)$.

With $0 = \log 1$ - proven by setting $b=0$, this relationship determines the logarithm.

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btw have you looked at this example? Intuitive proofs that $\lim\limits_{n\to\infty}\left(1+\frac xn\right)^n=e^x$