verify equicontinuous functions Suppose $f_n(x)=\sin(nx)$ on $[0.2\pi]$. Is the sequence of functions $\{f_n\}_{n \geq 1}$  equicontinuous on $[0.2\pi]$?
I try to compute derivative, which is $$f^{\prime}_n(x)=n\cos(nx)$$ which gives us $\sup_{x\in [0,2\pi]}|f^{\prime}_n(x)|=|n \cos(nx)|$, which is unbounded. So can we conclude that the sequence of functions is not equicontinous?
Remark: I don't know whether we can use derivative to show equicontinuity of a sequence of functions or not. I saw this somewhere else but I don't know why it works and why it doesn't.
EDIT: Is it always true that $|f_n(x) - f_n(y)| \leq \int_x^y{|f_n^{\prime}(t)|dt}$? So if $|f_n^{\prime}(t)|$ is bounded for all $n$, then the sequence of functions is equicontinuous?
 A: 
So can we conclude that the sequence of functions is not equicontinuous?

Not from the unboundedness of $f_n'$ alone. If we consider for example the family $\{ g_n \}$, where
$$g_n(x) = \sqrt{x+\frac{1}{n^2}},$$
we have $g_n'(0) = \frac{n}{2}$, but the family is equicontinuous nevertheless.

I don't know whether we can use derivative to show equicontinuity of a sequence of functions or not.

We can use the derivative to conclude the uniform equicontinuity if the family of derivatives is uniformly bounded, but the unboundedness of the family of derivatives is not sufficient to conclude that the original family is not equicontinuous.
For the family $f_n(x) = \sin (nx)$, we conclude that the family is not equicontinuous in $0$ by the fact that there are points arbitrarily close to $0$ where some $f_n$ attains the value $1$. To see that the family is not equicnotinuous in any point, a small modification of the argument is needed, given $\varepsilon > 0$, all $f_n$ for sufficiently large $n$ attain all values in $[-1,1]$ on every interval of length $\varepsilon$.
Concerning your

EDIT: Is it always true that $|f_n(x) - f_n(y)| \leq \int_x^y{|f_n^{\prime}(t)|dt}$? So if $|f_n^{\prime}(t)|$ is bounded for all $n$, then the sequence of functions is equicontinuous?

the estimate is correct (for $x \leqslant y$), and if we have a uniform bound $\lvert f_n'(t)\rvert \leqslant C$ for all $n$ and $t$, then the sequence is not only (uniformly) equicontinuous, it is even equilipschitz. The estimate yields the uniform equicontinuity of the sequence under the weaker assumption that the sequence of derivatives is uniformly integrable, that is, for every $\varepsilon > 0$ there is a $\delta > 0$ such that $\int_A \lvert f_n'(t)\rvert\,dt < \varepsilon$ for all (measurable) sets $A$ with measure $< \delta$. But since the uniform integrability of the derivatives is usually not easier to establish than the (uniform) equicontinuity of the family of functions, that criterion is of little practical value.
An occasionally useful criterion for equicontinuity that does not use the derivative (which may not exist) is that a (locally) uniformly convergent sequence of continuous functions is equicontinuous. The sequence is uniformly equicontinuous if the convergence is uniform and the limit function as well as all functions in the sequence are uniformly continuous. If the domain is compact, these latter conditions are implied by locally uniform convergence.
