Show that $(f_n)$ is equicontinuous, given uniform convergence
Solution 1:
You only need the uniform boundedness of the derivatives. Let $\epsilon > 0$, and choose $\delta = \epsilon/M$. Then for all $x<y \in [a,b]$ such that $|x-y| < \delta$, for every $n$, by the mean value theorem, there exists $c_n \in (x,y)$ such that $f_n(y) - f_n(x) = f_n'(c_n) \cdot (y - x)$. Therefore:
$$|f_n(y)-f_n(x)| \leq |f_n'(c_n)| \cdot |x - y| < M \cdot \epsilon/M = \epsilon$$
Solution 2:
Hint
$$ |f_n(x)-f_n(y)|=|(f_n(x)-f(x))+(f(x)-f(y))+(f(y)-f_n(y))| $$
$$ \leq |f_n(x)-f(x)|+|f(x)-f(y)|+|f(y)-f_n(y)| \,.$$
Now, use the assumptions you have been given.