Show that there is a sequence of functions continuously differentiable $f_{n}:[0,1] \to \mathbb{R}$, $n \in \mathbb{N}$, such that $f_{n}$ converges uniformly to $0$ but $f'_{n}$ not converges to $0$.
Take $$f_{n}(x) = \frac{x^{n+1}}{n+1}.$$ So, $$f'_{n}(x) = x^{n},$$ and $f'_{n} \to f$ pointwise where $f$ is the function $$f(x) = \begin{cases} 0,&\text{if}\;\;x \in [0,1)\\ 1,& \text{if}\;\; x=1.\end{cases}.$$ Is obvious that $f_{n} \to 0$ pointwise, but I couldn't show that the convergence is uniform.
Let $\epsilon>0$. There exists $N \in \mathbb{N}$ such that $\, 1/(n+1) < \epsilon \,$ for every $n \geq N$. Then for every such $n$ it follows that
$$ |f_{n}(x)-0| = \frac{|x|^{n+1}}{n+1} \leq \frac{1}{n+1} < \epsilon$$
for every $x \in [0,1]$: we have just verified the very definition of uniform convergence here.