An approach to the derivative of $e^x$ in Calculus 1 While I know there are plenty of topics on here concerning the derivative of $e^x$, I was wondering if what I have below is "okay" for the proof of it, or if there are holes in logic.
Proof:
$\displaystyle\frac{d}{dx}e^x = \lim_{h \to 0}{\frac{e^{x+h}-e^x}{h}}=\lim_{h\to 0}\frac{e^xe^h-e^x}{h}=\lim_{h \to 0} \frac{e^x(e^h-1)}{h}$
Now,
$\displaystyle e=\lim_{x \to \infty}(1+\frac{1}{x})^x$
Letting $\displaystyle x=\frac{1}{h} \Rightarrow e=\lim_{h \to 0}(1+h)^{1/h}$ because $h \to 0$ as $x\to \infty$.
If you raise both sides to the $h$ power you would then get $\displaystyle e^h=\lim_{h\to 0}(1+h)$.
You could then make a substitution for $e^h$ back into line one, though I am a little concerned. Namely, couldn't I simplify $\displaystyle \lim_{h\to 0}(1+h)=1$, but then substituting would not give the correct answer then.
NOTE: I wanted to find a proof that a Calculus 1 student could follow, but I also wanted to go about it as if we don't know the result of $\displaystyle\frac{d}{dx}\ln(x)$ yet nor have learned Taylor Series (as that would be Calculus 2).
 A: Here is not a proof, but another aspect may help students understand why the derivative holds this form, and it's actually a well-known explanation in some high-school textbooks with only elementary resources (I'm not saying this is 'good', but this is arguable).
This begins from opposite.
So, there is a kind of function, say $A(x)$, you don't know what it is, but you've known (or guess) the derivative of which must be itself.
$$\frac{\Delta A}{\Delta x}=A\quad\text{or}\quad\frac{\Delta A}{A}=\Delta x$$
(why here using $\Delta$? for the strict modern definition for limitation and derivative may be not introduced in high-school yet)
Simply, think about $A(0)=A_{0}=1$ for $x_{0}=0$, because we do not want to confused by the boundary in ODE problem these students completely have no idea at that time.
For this relationship should hold whatever the $x$ is, we set two sequences $A_{k}=A(x_{k})=A_{k-1}+\Delta A_{k-1}$ where $x_{k}=x_0+k\Delta x$ so you have
$$\frac{\Delta A_{k}}{A_{k}}=\Delta x$$
or
$$\frac{A_{k}+\Delta A_{k}}{A_{k}}=\frac{A_{k+1}}{A_{k}}=1+\Delta x$$
this hold for any $k$ as long as $\Delta x$ is small enough, for convenient, we choose an uniform step of $\Delta x$ which is $(x-x_{0})/N=x/N$ and $N$ could be infinity large. for any arbitrary $x=x_{N}$ and $A(x)=A_{N}$, you have
$$A_{N}=\frac{A_{N}}{A_{N-1}}\cdots\frac{A_{2}}{A_{1}}\frac{A_{1}}{A_{0}}=\left(1+\frac{x}{N}\right)^N$$
let $N\to\infty$ you will arrive $A(x)=e^x$.
actually, as we know, this is integration, literally anti-derivative, in regular calculus course.
A: A complementary answer to @Nanayajitzuki that justifies why taking $N\to\infty$ would give $A(x)=e^{x}$ (i.e., why define $e$ to do such a thing?).
Suppose we're so inclined to find the slope of the tangent line at $0$ of the function $f_{b}(x)=b^{x}$ for any base $b$. Using the definition of the derivative, we realize we're computing
$$
f_{b}'(0)=\lim_{h\to0}\frac{f(0+h)-f(0)}{h}=\lim_{h\to0}\frac{b^{h}-b^{0}}{h}=\lim_{h\to0}\frac{b^{h}-1}{h}.
$$
Suppose through experimentation we find that when $b=2$,
$$
f_{2}'(0)=\lim_{h\to0}\frac{2^{h}-1}{h}\approx0.693\ldots,
$$
and when $b=3$,
$$
f_{3}'(0)=\lim_{h\to0}\frac{3^{h}-1}{h}\approx1.099\ldots.
$$
With some justification that this family should be continuous in $b$, by the intermediate value theorem, there must exist some number $c$ such that $f_{c}'(0)=1$. Indeed, one may define $e$ to be the number such that $f_{e}'(0)=1$.
Then, one can quickly realize why this would be a fruitful thing to do (namely, it solves @Nanayajitzuki's boundary value problem):
$$
\frac{d}{dx}\left[e^{x}\right]=\lim_{h\to0}\frac{e^{x+h}-e^{x}}{h}=\lim_{h\to0}\frac{e^{x}e^{h}-e^{x}}{h}=e^{x}\lim_{h\to0}\frac{e^{h}-1}{h}=e^{x},
$$
hence we have the requisite construction.
