If $f_n $ converges uniformly to $f$, and the functions $f_n $ are differentiable, $f$ MIGHT NOT be differentiable, and $f_n'$ MIGHT NOT converge to $f'$.
$f_n : [-1, 1] \to \R $ defined by $f_n(x) := \sqrt[ ]{ \frac{ 1 }{ n^2 } + x^2 } $. These functions are differentiable. Also, we have that
$$\abs{ x } \leq f_n(x) \leq \abs{ x } + \frac{ 1 }{ n } $$
$\forall x \in [-1, 1]$. This can be proven by squaring both sides, Hence by squeeze test $f_n$ converges uniformly to absolute value function $f(x) = \abs{x}$. However, $\abs{ x } $ is not differentiable at $0$.
$f_n : [0, 2 \pi] \to \R , f_n(x) = n^{- 1/ 2} \sin (nx) $, and $f: [0, 2 \pi] \to \R, f(x) = 0$.
Since $sin(nx) \in [-1, 1]$, we have $d_\infty(f_n,f) \leq n ^{-1/ 2} $, where we use the uniform metric $d_\infty(f, g) = \supr_{x \in [0, 2\pi]}{ \abs{f(x)-g(x)}} $. Since $\limu{ n }{ \infty }n^{- 1/2} \to 0$.
Thus, $f_n $ converges uniformly to $f $.
However, $f'_n(x) = n^{1/ 2} \cos (nx) $, and so in particular $\abs{ f'_n(0) - f'(0)} = n^{1 / 2} $.
Thus $f'_n$ does not converge pointwise to $f'$, and so in particular does not converge uniformly either.
Here,
$$\der{ }{ x } \limu{ n }{ \infty } f_n(x) \neq \limu{ n }{ \infty } \der{ }{ x }f_n(x) $$
Let $[a, b]$ be an interval, and for every integer $n \geq 1$, let $f_n: [a, b] \to \R$ be a differentiable function whose derivative $f'_n: [a, b] \to \R$ is continuous. Suppose that the derivatives $f'_n$ converges uniformly to a function $g: [a, b] \to \R$. Suppose also that there exists a point $x_0$ such that the limit $\limu{ n }{ \infty } f_n(x_0)$ exists. Then t