Power Series is Termwise Differentiable within Radius of Convergence

Theorem
Let $\displaystyle f \left({x}\right) := \sum_{n \mathop = 0}^\infty a_n \left({x - \xi}\right)^n$ be a power series about a point $\xi$.

Let $R$ be the radius of convergence of $S$.

Then:
 * $\displaystyle \frac {\mathrm d}{\mathrm dx} f \left({x}\right) = \sum_{n \mathop = 1}^\infty \frac {\mathrm d}{\mathrm dx} a_n x^n = \sum_{n \mathop = 1}^\infty n a_n x^{n-1}$

Proof
Let $\rho \in \R$ such that $0 \le \rho < R$.

From Power Series Converges Uniformly within Radius of Convergence, $f \left({x}\right)$ is uniformly convergent on $\left\{{x: \left|{x - \xi}\right| \le \rho}\right\}$.

From Polynomial is Continuous, each of $f_n \left({x}\right) = a_n x^n$ is a continuous function of $x$.

From Power Rule for Derivatives:
 * $\dfrac {\mathrm d}{\mathrm dx} x^n = n x^{n-1}$

and again from Polynomial is Continuous, each of $\dfrac {\mathrm d}{\mathrm dx} f_n \left({x}\right) = n a_n x^{n-1}$ is a continuous function of $x$.

Then from Derivative of Uniformly Convergent Series of Continuously Differentiable Functions:
 * $\displaystyle \frac {\mathrm d}{\mathrm dx} f \left({x}\right) = \sum_{n \mathop = 1}^\infty \frac {\mathrm d}{\mathrm dx} a_n x^n = \sum_{n \mathop = 1}^\infty n a_n x^{n-1}$