[proofplan] The strategy uses the [Fundamental Theorem of Calculus](/theorems/632) to represent each $f_n$ as $f_n(c)$ plus an integral of $f_n'$. The [Interchange of Uniform Limits and Riemann Integrals](/theorems/259) converts [uniform convergence](/page/Uniform%20Convergence) of $f_n'$ into convergence of these integrals, yielding uniform convergence of $f_n$ to an explicitly identified [limit](/page/Limit). The [Uniform Limit Theorem](/theorems/258) gives continuity of the derivative limit $g$, and a second application of the Fundamental Theorem recovers $f' = g$. [/proofplan] [step:Represent each $f_n$ via the Fundamental Theorem of Calculus] Each $f_n$ is [continuously](/page/Continuity) [differentiable](/page/Derivative) on $[a,b]$, so $f_n'$ is continuous and therefore Riemann [integrable](/page/Integral) on every subinterval. The [Fundamental Theorem of Calculus](/theorems/632) (evaluation form) gives, for each $n \in \mathbb{N}$ and every $x \in [a,b]$: \begin{align*} f_n(x) &= f_n(c) + \int_c^x f_n'(t) \, dt. \end{align*} [/step] [step:Show the integral terms converge uniformly in $x$] For each $n \in \mathbb{N}$ and each $x \in [a,b]$: \begin{align*} \left| \int_c^x f_n'(t) \, dt - \int_c^x g(t) \, dt \right| &= \left| \int_c^x \bigl(f_n'(t) - g(t)\bigr) \, dt \right| \leq |x - c| \, \|f_n' - g\|_\infty \leq (b - a) \, \|f_n' - g\|_\infty. \end{align*} The right-hand side is independent of $x$ and tends to $0$ as $n \to \infty$ by [uniform convergence](/page/Uniform%20Convergence) $f_n' \to g$. Therefore the [sequence](/page/Sequence) of [functions](/page/Function) $x \mapsto \int_c^x f_n'(t) \, dt$ converges uniformly on $[a,b]$ to $x \mapsto \int_c^x g(t) \, dt$. [/step] [step:Establish uniform convergence of $f_n$ and identify the limit] Define $L := \lim_{n \to \infty} f_n(c) \in \mathbb{R}$ and $f: [a,b] \to \mathbb{R}$ by \begin{align*} f(x) &:= L + \int_c^x g(t) \, dt. \end{align*} By the integral representation, $f_n(x) = f_n(c) + \int_c^x f_n'(t) \, dt$. Therefore \begin{align*} \|f_n - f\|_\infty &= \sup_{x \in [a,b]} \left| f_n(c) - L + \int_c^x f_n'(t) \, dt - \int_c^x g(t) \, dt \right| \leq |f_n(c) - L| + (b - a)\|f_n' - g\|_\infty. \end{align*} Both terms tend to $0$ as $n \to \infty$ (the first by hypothesis, the second by [uniform convergence](/page/Uniform%20Convergence) of $f_n'$), so $f_n \to f$ uniformly on $[a,b]$. [/step] [step:Show $f$ is continuously differentiable with $f' = g$] Each $f_n'$ is [continuous](/page/Continuity) by hypothesis, and $f_n' \to g$ uniformly on $[a,b]$. By the [Uniform Limit Theorem](/theorems/258), $g$ is continuous on $[a,b]$. Since $g$ is continuous and $f(x) = L + \int_c^x g(t) \, dt$, the [Fundamental Theorem of Calculus](/theorems/632) (differentiation form) gives $f'(x) = g(x)$ for all $x \in [a,b]$. Continuity of $f' = g$ makes $f$ continuously [differentiable](/page/Derivative). [guided] We have established that $f(x) = L + \int_c^x g(t) \, dt$ where $g = \lim f_n'$ uniformly. To extract $f' = g$, we need two things: continuity of $g$, and the ability to differentiate the integral. First, $g$ must be continuous. Each $f_n'$ is continuous by hypothesis (each $f_n$ is continuously differentiable), and $f_n' \to g$ uniformly on $[a,b]$. By the [Uniform Limit Theorem](/theorems/258), the uniform limit of continuous functions is continuous. Therefore $g$ is continuous on $[a,b]$. Second, we differentiate under the integral sign. Since $g$ is continuous on $[a,b]$, the [Fundamental Theorem of Calculus](/theorems/632) (differentiation form) applies directly to the function $x \mapsto \int_c^x g(t) \, dt$. The conclusion is that this function is differentiable with derivative $g(x)$ at every $x \in [a,b]$. Combining: $f(x) = L + \int_c^x g(t) \, dt$ implies $f'(x) = 0 + g(x) = g(x)$ for all $x \in [a,b]$. Since $g$ is continuous, $f$ is continuously differentiable. This completes the proof: $(f_n)$ converges uniformly to $f$, and $f' = g = \lim_{n \to \infty} f_n'$. The argument used the Fundamental Theorem twice (once to represent $f_n$, once to differentiate $f$) and the Uniform Limit Theorem to establish continuity of $g$. [/guided] [/step]