Let $$f(x) = \sum_{k=0}^\infty a_kx^k$$ be a power series mapping reals to reals, with radius of convergence $R$. Suppose $f'(x_0)$ exists in $[-R,R]$ (take the one-sided limit if $x_0 = R$ or $x_0 = -R$). Then is it always true that $$f'(x_0) = \sum_{k=0}^\infty ka_kx^{k-1}?$$
This clearly holds if $x_0 \in (-R,R)$. But what about $x_0 = R$?
EDIT: Even though I have accepted the answer, I'm wondering if there still exists a counterexample if $a_k$ is restricted to the nonnegative reals for all $k$. It would be very helpful if someone can answer that.
No: consider $$ f(x) = \log{(1+x^2)}. $$ Then $$ f(x) = -\sum_{k=1}^{\infty} \frac{(-x^2)^k}{k}, $$ which is an alternating series with decreasing terms for $-1 \leqslant x \leqslant 1$, so it converges on $[-1,1]$. It does not converge outside this interval by applying the ratio test. Then $$ f'(x) = \frac{2x}{1+x^2} = \sum_{k=0}^{\infty} 2x(-x^2)^k $$ for $x \in (-1,1)$, and $f'(1)=1$, but the right-hand side evaluated at $x=1$ would be $$ 2\sum_{k=0}^{\infty} (-1)^k, $$ which is not Cauchy and so does not converge.