Need help with the proposition 4.2 in R. C. Churchill's paper on Liouville's Theorem in differential algebra I'm currently studying Liouville's theorem in differential algebra from this paper. I'm stuck on the proof of the propositions 4.2.
In the chapter 4, this "logarithmic derivative identity" is introduced:
$$\frac{\left(\prod_{j=1}^nt_j^{m_j}\right)^\prime}{\prod_{j=1}^nt_j^{m_j}}=\sum_{j=1}^nm_j\frac{t_j^\prime}{t_j}$$
According to the proposition 4.2, if the equation 
$$\alpha = \sum_{j=1}^mc_j\frac{\left(q_j(\ell)\right)^\prime}{q_j(\ell)}+\left(r(\ell)\right)^\prime$$
 holds, then the $q_j(t)$ are either non-constant monic irreducible polynomials in $K[\ell]$ or  elements of $K$.
$L=K(\ell)$ is a differential field extension, where $\ell$ is transcendental over $K$; $\alpha \in K$; $c_1,\ldots,c_m\in K_C$ are constants/elements of the kernel of the derivation; and $r(\ell), q_j(\ell)\in L$.
The proof says that the proposition follows from the logarithmic derivative identity, but I just can't see, how.
I did try writing the equation as
$$\alpha= \frac{\left(\prod_{j=1}^m\left(q_j(\ell)\right)^{c_j}\right)^\prime}{\prod_{j=1}^m\left(q_j(\ell)\right)^{c_j}} +\left(r(\ell)\right)^\prime$$
and substituting $q_j(\ell)=k_j\prod_{i=1}^{n_j}\left(q_{ji}(\ell)\right)^{n_{ji}}$, but that didn't seem to lead anywhere.
 A: As explained in the paper, we can first write each $q_j(\ell)$ as a quotient of two polynomials in $\ell$, which we can factorise into monic irreducible polynomials. Thus, we can write
$$q_j(\ell)=k_j\prod_{i=1}^{n_j}\left(q_{ji}\left(\ell\right)\right)^{n_{ji}}=k_j\cdot\hat{q}_j(\ell),$$
where $k_j\in K$, $q_{ji}(\ell)\in K[\ell]$, and $n_{ji}$ are integers not necessarily non-negative. Then,
$$\frac{(q_j(\ell))^\prime}{q_j(\ell)}=\frac{\left(k_j\hat{q}_j(\ell)\right)^\prime}{k_j\hat{q}_j(\ell)}=\frac{k_j^\prime\hat{q}_j(\ell)+k_j\hat{q}_j^\prime(\ell)}{k_j\hat{q}_j(\ell)}=\frac{k_j^\prime}{k_j}+\frac{\hat{q}_j^\prime(\ell)}{\hat{q}_j(\ell)}=\frac{k_j^\prime}{k_j}+\frac{\left(\prod_{i=1}^{n_j}\left(q_{ji}\left(\ell\right)\right)^{n_{ji}}\right)^\prime}{\prod_{i=1}^{n_j}\left(q_{ji}\left(\ell\right)\right)^{n_{ji}}}.$$
We can apply the logarithmic derivative identity on the last term:
$$\frac{(q_j(\ell))^\prime}{q_j(\ell)}=\frac{k_j^\prime}{k_j}+\sum_{i=1}^{n_j}n_{ji}\frac{\left(q_{ji}(\ell)\right)^\prime}{q_{ji}(\ell)}.$$
It is clear that when we multiply this by $c_j$ and sum over $j$, the sum is in the form we wanted.
