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

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

• Usually one defines $e^{x}$ and derives its properties before going to logarithms. So neither of these proof looks reasonable to me. – Kavi Rama Murthy May 12 at 23:20
• In What is Mathematics? the author defines logarithms first – George Dewhirst May 12 at 23:23

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.

• What if you start from the definition of e as that limit of $(1+1/n)^n$? I'm not the OP but I think that would be something interesting to include, maybe? – D.R. May 12 at 23:02
• Yeah that is a good point. Of course with the taylor series version it is easy to see it. This just leaves the $e^x = \lim (1+\frac{x}{n})^n$ – George Dewhirst May 12 at 23:05
• You cannot interchange a limit and a derivative without justification. – Kavi Rama Murthy May 12 at 23:14