Finding the derivative of a function using its series expansion

Question

I recently came across a proof which stated that the derivative of $\sin{x}$ is $\cos{x}$ because if we differentiate the Maclaurin series expansion of $\sin x$, we get the series expansion of $\cos x$. Is this a valid proof? My doubt is that the Maclaurin series expansion itself is defined based on the derivative of the function. The following expansion is the Maclaurin series where $a=0$ and $f^{(n)}$ denotes the $n^{th}$ derivative of the function. $$\sum_{n=0} ^ {\infty} \frac {f^{(n)}(a)}{n!} \, (x-a)^{n} $$ We can clearly see that we need the derivative in the first place before we can get the expansion. Is the proof valid?

Answer 1

You can do this:

Define

$e^x = \sum_\limits{n\to 0}^{\infty} \frac {x^n}{n!}$ and $e^{ix} = \cos x + i\sin x$

And then it will fall out that

$\sin x = \sum_\limits{n\to 0}^{\infty} \frac {x^{2n+1}}{(2n+1)!}$

and from here you can indeed show that $\frac {d}{dx} \sin x = \cos x$

However, using this approach we have not shown that this function has any of the properties we expect $\sin x$ to have, and that will have to be verified.