Let $\sigma(x)$ denote the sum of divisors of the positive integer $x$.
A number $y$ is said to be perfect if $\sigma(y)=2y$.
Denote the abundancy index of $z$ by $I(z)=\sigma(z)/z$.
Euler proved that an odd perfect number $N$, if one exists, must necessarily have the form $N = q^k n^2$ where $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
While considering the difference $$I(n^2) - I(q^k)$$ for $k=1$, I came across the interesting identity $$\frac{d}{dq}\bigg(I(n^2)-I(q)\bigg)=\frac{d}{dq}\bigg(\frac{q^2 - 2q - 1}{q(q+1)}\bigg)=\frac{3q^2 + 2q + 1}{q^2 (q+1)^2}.$$ This is interesting because of $$I(n^2)+I(q)=\frac{2q}{q+1}+\frac{q+1}{q}=\frac{3q^2 + 2q + 1}{q(q+1)}=q(q+1)\bigg(\frac{3q^2 + 2q + 1}{q^2 (q+1)^2}\bigg)$$ so that we have the identity (or differential equation (?)) $$q(q+1)\frac{d}{dq}\bigg(I(n^2)-I(q)\bigg)=I(n^2)+I(q).$$
Two questions:
[1] Is there a simple explanation for why the identity (or differential equation (?)) holds?
[2] Are there any other identities that could be derived in a similar fashion?
Update (July 25, 2020 - 10:15 AM Manila time)
I tried computing the derivative $$\frac{d}{dq}\bigg(I(n^2)+I(q)\bigg)$$ and I got $$q(q+1)\frac{d}{dq}\bigg(I(n^2)+I(q)\bigg)=q(q+1)\frac{d}{dq}\bigg(\frac{3q^2 + 2q + 1}{q(q+1)}\bigg)=q(q+1)\bigg(\frac{q^2 - 2q - 1}{q^2 (q+1)^2}\bigg)=I(n^2)-I(q).$$
There is nothing too special about the identities, as we shall soon see.
Hereinafter, we let $q^k n^2$ be an odd perfect number with special prime $q$, and we assume that the Descartes-Frenicle-Sorli Conjecture that $k=1$ holds true. Also, we denote the abundancy index of the positive integer $x$ by $I(x)=\sigma(x)/x$, where $\sigma(x)$ is the classical sum of divisors of $x$.
We compute:
$$\frac{d}{dq}(I(n^2) + I(q))=\frac{d}{dq}\bigg(\frac{3q^2 + 2q + 1}{q(q+1)}\bigg)=\frac{d}{dq}\bigg(3 - \frac{q - 1}{q(q + 1)}\bigg)=\frac{d}{dq}\bigg(\frac{-1}{q+1}+\frac{1}{q(q+1)}\bigg)$$ $$=\frac{d}{dq}\bigg(\frac{-2}{q+1}+\frac{1}{q}\bigg)=\frac{2}{(q+1)^2}-\frac{1}{q^2}= \frac{2q^2 - (q+1)^2}{(q(q+1))^2}$$
But $$I(n^2) - I(q) = \frac{2q}{q+1} - \frac{q+1}{q} = \frac{2q^2 - (q+1)^2}{q(q+1)}.$$
Similarly, we obtain:
$$\frac{d}{dq}(I(n^2) - I(q))=\frac{d}{dq}\bigg(\frac{q^2 - 2q - 1}{q(q+1)}\bigg)=\frac{d}{dq}(I(n^2) + I(q) - 2I(q))$$ $$=\frac{d}{dq}\bigg(3 - \frac{q - 1}{q(q + 1)} - \frac{2(q+1)}{q}\bigg)=\frac{d}{dq}\bigg(\frac{-1}{q+1}+\frac{1}{q(q+1)}+\frac{-2}{q}-2\bigg)$$ $$=\frac{d}{dq}\bigg(\frac{-2}{q+1}+\frac{1}{q}+\frac{-2}{q}-2\bigg)=\frac{d}{dq}\bigg(\frac{-2}{q+1}-\frac{1}{q}-2\bigg)=\frac{2}{(q+1)^2}+\frac{1}{q^2}$$ $$=\frac{2q^2 + (q+1)^2}{(q(q+1))^2}.$$
But $$I(n^2)+I(q)=\frac{2q}{q+1}+\frac{q+1}{q}=\frac{2q^2 + (q+1)^2}{q(q+1)}.$$
In the above computations, we have used the identity $$\frac{1}{q(q+1)}=\frac{1}{q}-\frac{1}{q+1}.$$