For a change, I will ask a derivative question. Please consider the Generalized W-Lambert/Product Logarithm function $\text W_k(z)$. Let’s see what happens when we try to differentiate with respect to the branch cut subscripted variable k using nth derivative notation. Note that partial derivative symbols are harder to type out:
Even though the function’s branch cuts are discrete, they can still use the discrete derivative. For a more generalized approach, it could be possible to just differentiate by ignoring the constraint of $k\in\Bbb Z$ and adding it back when done to see what happens. If the generalizations are the same regardless of the bolded link’s definitions, then we can ignore the constraint entirely.
$$\frac{d}{dk}\text W_k(z)=\text W^{(1,0)}_k(z), \frac{d}{dz} \text W_k^{(0,1)}(z)=\frac{\text W_k(z)}{x(\text W_k(z)+1)}=\frac{1}{e^{\text W_k(z)}+x} $$
Almost all special functions on Wolfram Mathworld has a closed form derivative, but the change of the branches of the Product Logarithm function are difficult to find. Here is an integral representation from the bolded link. Here assumes $k\in\Bbb Z, z\ne -\frac1e ,0$, but let’s see what happens if we differentiate with respect to k using the Special Case of the Leibniz Rule:
$$\frac{d}{dk}\text W_k(z) =\frac{d}{dk}\left(1+(\ln(z)+2\pi i k-1)e^{\frac{i}{2\pi}\int\limits_0^\infty\frac{1}{t+1}\ln\left(1-\frac{2\pi}{\ln(z)-\ln(t)+t+(2k+1)i\pi}\right)dt}\right)=(1-\ln(z))\frac{1}{\pi}e^{\frac{i}{2\pi}\int\limits_0^\infty\frac{1}{t+1}\ln\left(1-\frac{2\pi}{\ln(z)-\ln(t)+t+(2k+1)i\pi}\right)dt}\left[\int_0^\infty\frac{dt}{(t+1)(\ln(z)-\ln(t)+t+2ik-2\pi +i)}-\int_0^\infty\frac{dt}{(t+1)(\ln(z)-\ln(t)+t+2i k+i)}\right]+\frac d{dk}(2\pi i k)e^{\frac{i}{2\pi}\int\limits_0^\infty\frac{1}{t+1}\ln\left(1-\frac{2\pi}{\ln(z)-\ln(t)+t+(2k+1)i\pi}\right)dt}+2\pi i k\frac d{dk} e^{\frac{i}{2\pi}\int\limits_0^\infty\frac{1}{t+1}\ln\left(1-\frac{2\pi}{\ln(z)-\ln(t)+t+(2k+1)i\pi}\right)dt} $$
Now let’s finish the derivation by resubstitution and algebra with the next step already done:
$$ \frac{1-\ln(z)}{\pi}e^{\frac{i}{2\pi}\int\limits_0^\infty\frac{1}{t+1}\ln\left(1-\frac{2\pi}{\ln(z)-\ln(t)+t+(2k+1)i\pi)}\right)dt}\left[\int_0^\infty\frac{dt}{(t+1)(\ln(z)-\ln(t)+t+2ik-2\pi +i)}-\int_0^\infty\frac{dt}{(t+1)(\ln(z)-\ln(t)+t+2i k+i)}\right]+2\pi i e^{\frac{i}{2\pi}\int\limits_0^\infty\frac{1}{t+1}\ln\left(1-\frac{2\pi}{\ln(z)-\ln(t)+t+(2k+1)i\pi}\right)dt}- k e^{\frac{i}{2\pi}\int\limits_0^\infty\frac{1}{t+1}\ln\left(1-\frac{2\pi}{\ln(z)-\ln(t)+t+(2k+1)i\pi}\right)dt}\left[\int_0^\infty\frac{dt}{(t+1)(\ln(z)-\ln(t)+t+2ik-2\pi +i)}-\int_0^\infty\frac{dt}{(t+1)(\ln(z)-\ln(t)+t+2i k+i)}\right]= \left(\frac{1-\ln(z)}{\pi}-k\right)e^{\frac{i}{2\pi}\int\limits_0^\infty\frac{1}{t+1}\ln\left(1-\frac{2\pi}{\ln(z)-\ln(t)+t+(2k+1)i\pi}\right)dt}\left[\int_0^\infty\frac{dt}{(t+1)(\ln(z)-\ln(t)+t+2ik-2\pi +i}-\int_0^\infty\frac{dt}{(t+1)(\ln(z)-\ln(t)+t+2i k+i)}\right] +2\pi i e^{\frac{i}{2\pi}\int\limits_0^\infty\frac{1}{t+1}\ln\left(1-\frac{2\pi}{\ln(z)-\ln(t)+t+(2k+1)i\pi}\right)dt} = e^{\frac{i}{2\pi}\int\limits_0^\infty\frac{1}{t+1}\ln\left(1-\frac{2\pi}{\ln(z)-\ln(t)+t+(2k+1)i\pi)}\right)dt}\left[ \left(\frac{1-\ln(z)}{\pi}-k\right)\left[\int_0^\infty\frac{dt}{(t+1)(\ln(z)-\ln(t)+t+2ik-2\pi +i)}-\int_0^\infty\frac{dt}{(t+1)(\ln(z)-\ln(t)+t+2i k+i)}\right] +2\pi i \right]$$
Here is a partially closed form:
$$\text W_k(z) =1+(\ln(z)+2\pi i k-1)e^{\frac{i}{2\pi}\int\limits_0^\infty\frac{1}{t+1}\ln\left(1-\frac{2\pi}{\ln(z)-\ln(t)+t+(2k+1)i\pi)}\right)dt} \implies e^{\frac{i}{2\pi}\int\limits_0^\infty\frac{1}{t+1}\ln\left(1-\frac{2\pi}{\ln(z)-\ln(t)+t+(2k+1)i\pi)}\right)dt}= \frac{\text W_k(z) -1}{\ln(z)+2\pi i k-1}\implies e^{\frac{i}{2\pi}\int\limits_0^\infty\frac{1}{t+1}\ln\left(1-\frac{2\pi}{\ln(z)-\ln(t)+t+(2k+1)i\pi)}\right)dt}\left[ \left(\frac{1-\ln(z)}{\pi}-k\right)\left[\int_0^\infty\frac{dt}{(t+1)(\ln(z)-\ln(t)+t+2ik-2\pi +i)}-\int_0^\infty\frac{dt}{(t+1)(\ln(z)-\ln(t)+t+2i k+i)}\right] +2\pi i \right] = \frac{2\pi \text W_k(z) -1}{\ln(z)+2\pi i k-1}\left[ \left(\frac{1-\ln(z)}{\pi}-k\right)\int_0^\infty \frac{dt}{(t+1)(\ln(z)-\ln(t)+t+(2k+1)i)(\ln(z)-\ln(t)+t+(2k+1)i-2\pi)}+ i \right]=\frac d{dk} \text W_k(z), k\text{ is any constant}$$
Note that the restrictions on the branch cut k can be adjusted.
Here is a sum representation using the bolded link using the Stirling Numbers of the first kind $S^{(x)}_y$:
$$\frac{d}{dk}\text W_k(z)= \frac{d}{dk}\left((2i\pi k+\ln(z))-\ln(2i\pi k+\ln(z))-\sum_{p=0}^\infty\frac{(-1)^p}{(2i\pi k +\ln(z))^p}\sum_{j=1}^p\frac{S^{(p-j+1)}_p \ln^j(2i\pi k+\ln(z))}{j!} \right) , z\to 0,\infty=2\pi i+\frac{2i\pi}{\ln(z)+2i\pi k}-2 i \pi\sum_{p=0}^\infty \sum_{j=1}^p \frac{(-1)^p S^{(p-j+1)}_p\ \ln^{j-1}(2 i k \pi + \ln(z)) (j - p \ln(2 i k π + \ln(z))) }{(i (2 k \pi - i \ln(z)))^{p+1} j!}=? $$
There area few other representations, but only one should work. Also, the branches themselves have a distinct pattern. Please also see details in the Elementary properties, branches and range section. Note there could be typos. What is a closed form of derivative of the Lambert-W function with respect to the function’s branches? I have given a few possible forms here. Please correct me and give me feedback!
Just for fun, the Wright Omega function can also have the interesting identity that:
$$ω(z)\mathop=^\text{def} \text W_{\left \lceil\frac{\text{Im}(z)}{2\pi}-\frac12\right\rceil}\left(e^z\right)\implies \text W_k(z)=ω(\ln(z)+2i\pi k)$$
which can only be used for the discrete version because it causes the branch cut to be discrete requiring discrete calculus for first solution, ignoring the differential equation for the function in the link:
Here is a discrete calculus solution using the difference quotient. The integer nature of the branch cuts creates the smallest possible value of $|h|=1$ just to have 2 variations:
$$\frac{\Delta \text W_k(z)}{\Delta k}= \frac{\Delta }{\Delta k}\text W_k(z)= \lim_{h\to 0}\frac{\text W_{k+h}(z)-\text W_k(z)}{h} \mathop=^\text{discrete} \pm \text W_{k\pm1}(z)\mp \text W_k(z) =\text W_{k+1}(z)- \text W_k(z), \text W_k(z)- \text W_{k-1}(z) $$
Here is a more general equation using Wolfram Functions $1.32.27.3.1$ and logarithmic differentiation:
$$\ln\left(\text W_k(z)\right)=\ln(z)-\text W_k(z)+2i\pi k\implies \frac{d}{dk} \ln\left(\text W_k(z)\right)=\frac{d}{dk}\left(\ln(z)-\text W_k(z)+2i\pi k \right)\implies \frac{\text W^{(1,0)}_k(z)}{\text W_k(z)}=0-\text W^{(1,0)}_k(z) +2\pi i=2\pi i -\text W^{(1,0)}_k(z) \implies 2\pi i= \frac{\text W^{(1,0)}_k(z)}{\text W_k(z)} +\text W^{(1,0)}_k(z) = \text W^{(1,0)}_k(z)\left(\frac1{\text W_k(z)}+1\right)\implies \text W^{(1,0)}_k(z)=\frac{d}{dk} \text W_k(z)=\frac{2i\pi}{\frac1{\text W_k(z)}+1}=\frac{2i\pi \text W_k(z)}{\text W_k(z)+1}$$
Similarly we can solve for $\text W_k(z)$:
$$\ln\left(\text W_k(z)\right)=\ln(z)-\text W_k(z)+2i\pi k\implies \text W_k(z)= \text W_b\left(e^{2i\pi k}z\right) = \text W\left(e^{2i\pi k}z\right)$$
where $b$ just adds more branches of the function.
This means that $\text W_k(z)$ satisfies the following first order nonlinear ordinary differential equation. I assume that the constant of integration $c$ is $z$:
$$y’(k)=\frac{2i\pi\, y(k)}{y(k)+1}\implies y(k)= \text W_k(z) = \text W\left(e^{c+2i\pi k}\right)=\text W_k\left(ce^{2\pi i k}\right) =\text W_k\left(ze^{2\pi ik}\right) $$
I am still unsure about the differential equation result generalization.
Look at the $\frac{d\,w(k,z)}{dk}$ result for proof. Please correct me and give me feedback!