[proofplan] The argument has two parts. First, we show the derived [series](/page/Series) has the same [radius of convergence](/theorems/273) $R$ using the root test: $\limsup |nc_n|^{1/n} = \limsup |c_n|^{1/n}$ since $n^{1/n} \to 1$. Second, we apply the termwise [differentiability](/page/Derivative) theorem on each compact sub-interval $[a - r, a + r]$ with $r < R$, using [uniform convergence](/page/Uniform%20Convergence) of the derived series from [Theorem 273](/theorems/273) and convergence at $x = a$. [/proofplan] [step:Show the derived series has radius of convergence $R$] The Cauchy--Hadamard formula gives $1/R' = \limsup_{n \to \infty} |n c_n|^{1/n}$ for the derived [series](/page/Series). Since $n^{1/n} \to 1$ as $n \to \infty$, \begin{align*} \limsup_{n \to \infty} |n c_n|^{1/n} &= \limsup_{n \to \infty} n^{1/n} |c_n|^{1/n} = 1 \cdot \limsup_{n \to \infty} |c_n|^{1/n} = \frac{1}{R}. \end{align*} Therefore $R' = R$. [/step] [step:Verify the hypotheses for termwise differentiation on each compact sub-interval] Fix $x \in (a - R, a + R)$ and choose $r$ with $|x - a| < r < R$. On the compact interval $[a - r, a + r]$, the original [series](/page/Series) converges (in particular at $x$), and the derived series $\sum n c_n(t - a)^{n-1}$ [converges uniformly](/page/Uniform%20Convergence) by [Theorem 273](/theorems/273) (part 3) applied to the derived series, which has the same radius $R > r$. Each term $c_n(t - a)^n$ is [continuously](/page/Continuity) [differentiable](/page/Derivative) on $[a - r, a + r]$. The termwise differentiability theorem for series therefore applies, yielding \begin{align*} f'(x) &= \sum_{n=1}^\infty n c_n(x - a)^{n-1} \end{align*} on $(a - r, a + r)$. Since $r < R$ was arbitrary (subject to $|x - a| < r$), this holds for all $x \in (a - R, a + R)$. [guided] We want to differentiate the [power series](/page/Power%20Series) $f(x) = \sum_{n=0}^\infty c_n(x - a)^n$ term by term. The termwise differentiability theorem (for series of [functions](/page/Function)) requires two conditions on each compact sub-interval $[a - r, a + r]$ with $r < R$: (i) The series of derivatives $\sum n c_n(t - a)^{n-1}$ converges uniformly on $[a - r, a + r]$. Since the derived series has radius $R$ (by the previous step) and $r < R$, [Theorem 273](/theorems/273) (part 3) gives [uniform convergence](/page/Uniform%20Convergence) on $[a - r, a + r]$. (ii) The original series converges at some point. It converges at $x = a$ (giving $f(a) = c_0$). With both conditions verified, the theorem gives \begin{align*} f'(x) &= \sum_{n=1}^\infty n c_n(x - a)^{n-1} \end{align*} on $(a - r, a + r)$. Every $x \in (a - R, a + R)$ lies in some such interval, so the formula holds on all of $(a - R, a + R)$. [/guided] [/step]