Let $f\left(x\right)=x^2+1$ on the interval $\left[-\pi,\pi\right]$, which is extended periodically to $\mathbb{R}$.
I have calculated the Fourier series of $f$ to be $$f\left(x\right)=\dfrac{\pi^2+3}{3}+4\sum^{\infty}_{n=1}\dfrac{\left(-1\right)^n}{n^2}\cos\left(nx\right)$$
Considering the derivative of $f$, write down the Fourier series of the function $g\left(x\right)=x$ on $\left(-\pi,\pi\right]$, again extended periodically to $\mathbb{R}$, in the form $$g\left(x\right)=c_0+\sum^{\infty}_{n=1}c_n\cos\left(nx\right)+d_n\sin\left(nx\right)$$
I see that the derivative of $f$ is $2x$, but I am not sure of how to do this without calculating explicitly the coefficients.
$f'(x) = 4\sum_{n = 1}^{\infty} \dfrac{(-1)^n}{n^2}\dfrac{d}{dx}\cos(nx)$
$2x = 4\sum_{n = 1}^{\infty} \dfrac{(-1)^n}{n^2}(-n)\sin(nx)$
$x = 2\sum_{n = 1}^{\infty} \dfrac{(-1)^{n+1}}{n}\sin(nx)$
Then, $c_n = 0$ for $n \geq 0$ and $d_n = 2\dfrac{(-1)^{n+1}}{n}$.