Suppose we have a function $f(t)$ which is periodic, continuous and differentiable everywhere. Call its period $T_0$. Suppose that it can be written as $$f(t) = \sum_{k = -\infty}^\infty a_k e^{j\left(\frac{2\pi}{T_0}\right)kt}$$ where $$a_k = \frac{1}{T_0}\int_{t=0}^{T_0}f(t)e^{-j\left(\frac{2\pi}{T_0}\right)kt}dt$$
I was asked to determine the sequence $\{b_n\}$ such that $b_n$ is the $n^\text{th}$ coefficient of the fourier series of $\frac{d}{dt}f(t)$.
What I did was the following:
In the equation of $a_k$, I changed $f(t)$ with $\frac{d}{dt}f(t)$, which gave me an integral that I could solve with integration by parts: $$b_k = \frac{1}{T_0}\int_{t=0}^{T_0}e^{-j\left(\frac{2\pi}{T_0}\right)kt}d(f(t))$$ In the result, I got that $b_k = 2\pi j \frac{k}{T_0^2} a_k$, but I cannot be sure if I got an extra $T_0$ or missed something somewhere else. Squaring the period didn't seem that cute and I saw a few posts similar to this that had answers not that similar to what I found.
Am I missing anything?
Let $$ f\left(t\right)=\sum_{n=-\infty}^{\infty}\hat{f}\left(n\right)e^{\frac{2\pi j}{T}nt}\Rightarrow\frac{d}{dt}f\left(t\right)=\frac{d}{dt}\sum_{n=-\infty}^{\infty}\hat{f}\left(n\right)e^{\frac{2\pi j}{T}nt}=\sum_{n=-\infty}^{\infty}\hat{f}\left(n\right)\frac{d}{dt}e^{\frac{2\pi j}{T}nt}= \\ \sum_{n=-\infty}^{\infty}\hat{f}\left(n\right)e^{\frac{2\pi j}{T}nt}\cdot\frac{2\pi j}{T}n=\sum_{n=-\infty}^{\infty}\hat{g}\left(n\right)e^{\frac{2\pi j}{T}nt} $$ Where $$ \hat{f}\left(n\right)=\frac{1}{T}\int\limits _{-\frac{T}{2}}^{\frac{T}{2}}f\left(t\right)e^{-\frac{2\pi j}{T}nt}dt,\hat{g}\left(n\right)=\hat{f}\left(n\right)\cdot\frac{2\pi j}{T}n $$