An approach to the derivative of $e^x$ in Calculus 1

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.

$$frac{Delta A}{Delta x}=Aquadtext{or}quadfrac{Delta A}{A}=Delta x$$

(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 $x_{k}=x_0+kDelta x$ 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

$$A_{N}=frac{A_{N}}{A_{N-1}}cdotsfrac{A_{2}}{A_{1}}frac{A_{1}}{A_{0}}=left(1+frac{x}{N}right)^N$$

let $Ntoinfty$ you will arrive $A(x)=e^x$.

actually, as we know, this is integration, literally anti-derivative, in regular calculus course.

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