If $R(z)$ is a rational function of order $n$, how large and how small can the order of $R'(z)$ be?
I am not sure whether my idea is right or wrong:
$R(z)=\frac{P(z)}{Q(z)}$, if deg$P(z)=n$, deg$Q(z)=m\le n$. $R'(z)=\frac{P'(z)Q(z)-Q'(z)P(z)}{Q(z)^2}$, deg$P'(z)Q(z)-Q'(z)P(z)\le m+n-1$, deg$Q(z)^2=2m$. To get the largest, take $m=n$, then the order of $R'(z)=2n$, to get the smallest, $P'(z)Q(z)-Q'(z)P(z)$ may cancel out to leave a constant. the order of $R'(z)=2m$.
We can consider deg$Q(z)=n$, deg$P(z)=m\le n$ similarly.
So the order of $2\le 2m\le R'(z)\le 2n$, is it right?
You have not considered that your expression for $R'$ can be simplified, which lowers its degree, and also you haven't dealt with the cases where $\deg Q = n$ and $\deg P = m < n$.
A rational fraction of degree $n$ has $n$ poles when counted with multiplicity, so let us see if there are nice relations between poles of $f$ and poles of $f'$ :
If $z \in \Bbb C$ is a pole of $f$ of order $k > 1$, then $f'$ has a pole of order $k+1$ at $z$, and there is no other way for $f'$ to have a pole there.
If $\infty$ is a pole of $f$ or order $k > 1$ (meaning $\deg P = k + \deg Q$), then $f'$ has a pole of order $k-1$ at $\infty$, and again there is no other way to get a pole for $f'$ at $\infty$.
So counting up the number of poles of $f'$ with multiplicity, you get that $\deg f' = \deg f +$ number of distinct finite poles - number of distinct infinite poles.
Since $f$ has $\deg f$ poles in total, if $\deg f > 2$ you can have $\deg f'$ anywhere between $\deg f - 1$ (when $f$ has all its poles at $\infty$, so when it is a polynomial) and $2\deg f$ (when $f$ has all its poles distinct and finite)
If $\deg f = 1$ then $\deg f' = 0$ if the pole is at $\infty$ and $2$ if it's anywhere else (and $1$ is not possible to get).