Is it true that if $f_{n}\rightarrow f$ uniformly converges then $f^{\prime}_{n}\rightarrow f^{\prime}$? Let $f_{n}$ be some sequence of functions.
If $f_{n}$ uniformly converges to $f$  then is it true that  $f^{\prime}_{n}\rightarrow f^{\prime}$?
Is there an example that proves/disproves this?
 A: Counterexample: $f_n = \frac1n\sin(n^2 x)$.
A: This is false in quite a strong sense. The Weierstrass function is a uniform limit of differentiable functions, but is nowhere differentiable. By the Stone—Weierstrass Theorem, any continuous function on a closed bounded interval is a uniform limit of polynomials.
A: The additional condition required is that a function $g$ exists to which the $f'_n$ uniformly converge (in which case of course $g=f'$). In particular, this PDF proves on p. 9 that this condition suffices, but on p. 8 it shows that $f_n=\frac{x}{1+nx^2}$ is a uniformly convergent sequence whose derivatives form a convergent but not uniformly convergent counterexample.
A: As mentioned above this not true in general. But one might still aks what a sufficient condition for "$f_n' \to f'$ uniformly" might be. If you already know that $f_n$ converges uniformly to $f$ and $f_n'$ converges uniformly to some function say $g$, then by the fundamental theorem of calculus one indeed obtains: $g=f'$ hence $\space$$f_n' \to f'$ uniformly.
