Let $(S,+,\cdot)$ be a semiring; a derivation on $S$ is a map $\partial : S \to S$ that is linear and Leibniz, in the sense that
Now, assume that $g\in S$ has a multiplicative inverse; what is the derivative of $g^{-1}$, considering that $-\frac{\partial g}{g^2}$ might not exist?
This is not an answer, but is too long for a comment.
It was already mentioned in the comments under the OP that, if $\partial$ is a derivation on a (commutative or non-commutative) semiring $S$ with $\partial 1_S = 0_S$, then the "usual formula" for the derivative of a multiplicative unit $u$ carries over as expected; namely, $\partial u$ is an additive unit and we have $\partial u^{-1} = -u^{-1} \cdot \partial u \cdot u^{-1}$. Otherwise, $\partial 1_S$ is still an idempotent element of the additive monoid of $S$, which motivates the following observations:
(i) If $\partial$ is a derivation on a commutative semiring $S$ and $e$ is an idempotent of the additive monoid of $S$, then the function $\partial_e \colon S \to S \colon x \mapsto \partial x + ex$ is also a derivation on $S$. In particular, note that
$$ \partial_e(xy) = (x\partial y + y \partial x) + (xye + xye) = x (\partial y + ye) + y (\partial x + xe) = x \partial_e y + y \partial_e x, $$ for all $x, y \in S$; that is, $\partial_e$ is Leibniz.
(ii) Let $S$ be an idempotent commutative semiring ("idempotent" means that $x = x+x$ for every $x \in S$). It is easily seen that the function $\partial_e \colon S \to S \colon x \mapsto xe$ is a derivation on $S$ for each $e \in S$; this is, in fact, a special case of the construction considered under item (i), with $\partial$ the trivial derivation on $S$. Moreover, $\partial_e u^{-1} = u^{-1} e = u^{-2} \partial_e u$ for every multiplicative unit $u \in S$ (this looks similar to the "usual formula" for the derivative of a multiplicative unit: I don't like to have a good explanation for the "coincidence", but I don't). Note, though, that the only unit of $(S, +)$ is $0_S$.
(iii) For a concrete example, let $S$ be the min-plus semiring $(\mathbf R \cup \{+\infty\}, \oplus, \otimes)$, where the operations are defined by $x \oplus y := \min(x,y)$ and $x \otimes y := x + y$ for all $x, y \in \mathbf R \cup \{+\infty\}$. This is an idempotent commutative semiring whose multiplicative identity is $0$ and whose group of (multiplicative) units is $\mathbf R$. So, the construction considered under item (ii) is not vacuous.
(iv) The set $\mathcal D(S)$ of all derivations on a (commutative or non-commutative) semiring $S$ is naturally made into a left/right $S$-semimodule, by endowing it with the operation of pointwise addition and scalar multiplication induced by $S$ (the additive identity is the trivial derivation). Loosely speaking, this means that also the formulas for the derivative of a multiplicative unit of $S$ will form a left/right $S$-semimodule.
(v) One can always extend the operations of a semiring $S$ by adding a "point at infinity" $\infty$ and taking, for every $x \in S \cup \{\infty\}$, $$ x + \infty = \infty + x := \infty \quad\mathrm{and}\quad x \cdot \infty = \infty \cdot x := \left\{ \begin{array}{ll} 0_S & \mathrm{if } x = 0_S, \\ \infty & \mathrm{if } x \ne 0_S. \end{array} \right. $$ It is readily checked that $(S \cup \{\infty\}, +, \cdot\,)$ is a semiring (with the same additive and multiplicative units of $S$) whose additive monoid is annihilated by $\infty$. Moreover, if the only unit of the additive monoid of $S$ is the identity $0_S$, then the function $$ \partial \colon S \cup \{\infty\} \to S \cup \{\infty\} \colon x \mapsto \left\{ \begin{array}{ll} 0_S & \mathrm{if } x = 0_S, \\ \infty & \mathrm{if } x \ne 0_S \end{array} \right. $$ is a derivation on the extended semiring. On the one hand, I take this as a sign that something is "wrong" from a conceptual point of view: Maybe the condition $\partial 1_S = 0_S$ should be part of the definition of a derivation (the condition is implied by Leibniz's law when it comes to rings, since the additive monoid of a ring is a group, hence cancellative). On the other hand, it is curious that $\partial u^{-1} = u^{-2} \partial u$ for every multiplicative unit $u \in S$ (cf. item (ii)), isn't it?
