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?
Comment by Paul Sinclair: $I$ is defined on a discrete set, the natural numbers. Differentiation requires functions defined on a continuum. How are you extending $I$ to a continuum so that you can perform this differentiation?
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
$$q(q+1)\frac{d}{dq}\bigg(I(n^2)+I(q)\bigg)=q(q+1)\frac{d}{dq}\bigg(\frac{3q^2 + 2q + 1}{q(q+1)}\bigg)=q(q+1)\bigg(\frac{q^2 - 2q - 1}{q^2 (q+1)^2}\bigg)=I(n^2)-I(q).$$
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
$$q(q+1)\frac{d}{dq}\bigg(I(n^2)+I(q)\bigg)=q(q+1)\frac{d}{dq}\bigg(\frac{3q^2 + 2q + 1}{q(q+1)}\bigg)=q(q+1)\bigg(\frac{q^2 - 2q - 1}{q^2 (q+1)^2}\bigg)=I(n^2)-I(q).$$
