Finding the derivative of a function using its series expansion

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?

• So how do we find the McLaurin series of $sinx$ in the first place? – imranfat Jan 31 '18 at 1:44
• That is exactly my question @imranfat – Vishal Subramanyam Jan 31 '18 at 1:46
• The question goes much deeper then you think. Conventionally, the MacLaurin series is set of based on multiple derivatives and traditionally that is how we establish Series of a host of transcendental functions. Alternatively, one can define the MacLaurin series for the sine and work backwards to establish its properties and so forth...This is less intuitive, but it can be done – imranfat Jan 31 '18 at 1:49
• @imranfat Could you explain a little more? – Vishal Subramanyam Jan 31 '18 at 1:50
• Have you taken a course on Complex analysis? – imranfat Jan 31 '18 at 1:51

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.

• might be worth noting why one can differentiate termwise or that doing so is not always ok. +1 regardless – qbert Jan 31 '18 at 2:03