# 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.