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What does one have to prove/show in order to justify $$ f(x)=\sum_{n=1}^k c_n \sin(nx) $$ has derivative $$ f'(x)=\sum_{n=1}^k c_n n \cos(nx) ?$$

I am used to assuming this to be true. Say also that $\sum\limits_1^\infty |c_n| < \infty$. Thank you.

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    For finite sums there is no problem (since you can "expand it out and do term by term"). So the problem is that that (infinite) trigonometric series $\sum c_n \sin(nx)$ and $\sum c_n n \cos(nx)$ may have convergence issues. If $c_n$ is merely absolutely summable, the infinite series $\sum c_n n\cos(nx)$ may not converge!2011-09-19
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    In the case $k = \infty,$ a sufficient condition (and one that you probabaly already know) is that the sums $\sum c_n \mathrm{sin}(nx), \sum c_nn\mathrm{cos}(nx)$ converge uniformly on compact sets.2011-09-19

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