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!
