Prove that the derivative of exponential is itself without using derivative

Question

It's a little problem that I found interesting . To solve it without using differentiation we solve the following equation : Let $x,y$ be real numbers then solve : $$e^x=\frac{e^x-e^y}{x-y}$$ We multiply by $(x-y)$ it gives : $$e^x(x-y)=e^x-e^y$$ Or: $$e^x(x-y-1)=-e^y$$ We make the following substitution $u=x-y-1$

We have : $$e^{u+y+1}(u)=-e^y$$ Or $$ue^u=-e^{-1}$$ We introduce the product log function it gives : $$W(-e^{-1})=u$$ Or : $$u=-1$$ So we have $x=y$ and conclude that the derivative of exponential is itself .

My question :

Have you a similar method wich doesn't use derivative .

Thanks a lot for your interest

Answer 1

Perhaps, you will not like the following solution because I will use differentiation (but not in the case of $e^x$.)

So I try to find a function whose derivative is itself. I cannot be this smart but assume that I try the following. What about $$f_1(x)=1+x?$$ The derivative is $1$. So my first choice is wrong. But, "I have an idea". Let $$f_2(x)=1+x+\frac12 x^2.$$ The derivative is wrong again but I have a more general idea now. What about $$f_n(x)=1+x+\frac12x^2+\frac1{3!}x^3+\dots+\frac1{n!}x^n.$$

The derivative is still wrong but promising:

$$f_n(x)=1+x+\frac12x^2+\frac1{3!}x^3+\dots+\frac1{(n-1)!}x^{n-1}.$$

So, I expect that the function which I define the following way will be OK:

$$f_n(x)=1+x+\frac12x^2+\frac1{3!}x^3+\dots+\frac1{n!}x^{n}+\dots$$

Let's call the limit of the series above (if it exists):

$$e^x.$$

This "imaginary" function is, hopefully a function whose derivative is itself.

Answer 2

If you mean to prove that $(e^x)'=e^x$ without using the definition of the derivative (directly), then you could do it as follows:

Let $u=\ln(x)$, then $$e^u=x$$ differentiating both sides, and using the chain rule, we get

$$(e^u)'u'=1\\(e^u)'\frac{1}{x}=1\\(e^u)'=x\\\boxed{(e^u)'=e^u}$$