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My friend was asking me how to find derivative of $e^x$ using first principle of derivative. I gave him the following method: $$f(x) = e^x$$

$$f'(x) = \lim_{h\to0}\dfrac{e^{x+h}-x}{h}$$ $$f'(x) = \lim_{h\to0}\dfrac{e^x(e^{h}-1)}{h}$$ Now using the well known limit $\lim_{x\to0}\dfrac{e^x- 1}{x}$ $$f'(x) =e^x(1) => e^x$$

Then he asked me for the proof of $\lim_{x\to0}\dfrac{e^x- 1}{x}$. $$e^x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + ...$$ $$e^x - 1= x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + ...$$ $$\dfrac{e^x-1}x = 1 + \dfrac{x}{2!} + \dfrac{x^2}{3!} + ...$$ $$\lim_{x\to0}\dfrac{e^x-1}x = \lim_{x\to0}1 + \dfrac{x}{2!} + \dfrac{x^2}{3!} + ...$$ $$\lim_{x\to0}\dfrac{e^x-1}x = 1$$

He told me that for finding derivative of $e^x$ we use Maclaurin series and for finding Maclaurin series, we use derivative. So they are inter related. How can I satisfy him?

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    You need to say how you are defining $\exp(x)$. If you define it as that power series (and then demonstrate all the necessary properties), then it is not circular. It's clear that the formal derivative of that power series is itself, for instance. – lulu Mar 17 '22 at 13:11
  • Depends on what definition of e you take. You could consider it to be the function such that f(0)=1 and it is its own derivative. There are many other definitions like the limit of product one which can be found on site. And nice that you went all the way to ask over here for your friend lol – tryst with freedom Mar 17 '22 at 13:11

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