needing help for proofing $\frac{de^x}{dx}=e^x$

Question

could anybody explain why do we proof $\frac{de^x}{dx}=e^x$ in this way "we know $y=e^x=f(x)$ and $(f(y)^-1)' =\frac{1}{f'(x)}$ ,so $\frac{dy}{dx}=\frac{de^x}{dx}=\frac{dln^-1}{dx}=\frac{df(x)^-1}{dx}=\frac{1}{f'(f(x)^-1)}=\frac{1}{f'(e^x)}\Rightarrow\frac{1}{\frac{1}{e^x}}=e^x$" instead of "$y=e^x\Rightarrow lny=x\Rightarrow\frac{1}{y}\frac{dy}{dx}=1\Rightarrow\frac{dy}{dx}=y\Rightarrow\frac{de^x}{dx}=e^x $" thank you for any help

Answer 1

It depends which definition of $e^x$ we use. Indeed, some definitions say that $e^x$ is precisely the solution to the ODE $\frac{dy}{dx}=y$.

You might have started from the idea that $\ln(x)$ is a function such that $\frac{d}{dx}\ln(x) = \frac{1}{x}$. Then you can use inverse function theorem as is done here.

It turns out all of the definitions of $e^x$ are equivalent anyway.

Notice if we go from $e^x = \lim (1+\frac{x}{n})^n$,

$\frac{d}{dx}e^x = \lim (1+\frac{x}{n})^{n-1} = \lim (1+\frac{x}{n})^{n}/\lim(1+\frac{x}{n}) = e^x$.

Here, we can see that the two limits converge so can divide them out.

We exchanged the limit and derivative. Let's justify that. The limit $\lim (1+\frac{x}{n})^n$ converges uniformly in $x$ to $e^x$, hence we can exchange them.