$\sin{x}$ is convergent to its power series for all $x$. I proved this using the Lagrange formula for the remainder showing $R_n(x) \to 0$ as $n \to \infty$. Does this imply the power series of $\sin x$ uniformly converges? Because next I'm trying to interchange $\tfrac{d}{dx}$ and summation which is allowed if the sum converges uniformly:
$\dfrac{d}{dx} \sin{x} = \dfrac{d}{dx}\sum_{n=0}^\infty (-1)^{n} \, \dfrac{x^{2n + 1}}{(2n +1)!} = \sum_{n=0}^\infty (-1)^{n} \, \dfrac{d}{dx} \dfrac{x^{2n + 1}}{(2n +1)!}$