The following section in Kontsevich' and Soibelman's paper I don't understand:
Let $F$ be a polynomial functor [this is the author's terminology for analytic functor as explained here], $P= Free_{\mathcal{OP}}(F)$ be the corresponding free operad. Let $g_p$ be the Lie algebra (in the symmetric monoidal category $\mathcal{C}$) of derivations of the operad $P$.
What is a derivation of the operad $P$? Is it a morphism of operads $d:P\rightarrow P$ such that $d(f\circ_i g)=f\circ_i d(g)+d(f)\circ_i g$ for all $f\in P(n), g\in P(m)$? References welcome.
A derivation $d : P \to P$ is a sequence of $\Sigma_n$-equivariant maps $d : P(n) \to P(n)$ such that, for homogenous elements $a \in P(n)$, $b_1 \in P(r_1)$, ..., $b_n \in P(r_n)$, one has: $$d\bigl(a(b_1, \dots, b_n)) = (d(a))(b_1, \dots, b_n) + \sum_{i=1}^n (-1)^{\epsilon_i} a(b_1, \dots, d(b_i), \dots, b_n),$$ where $\epsilon_i = |a| + |b_1| + \dots + |b_{i-1}|$.
This is defined e.g. in Section 6.3.1 of Loday-Vallette' book Algebraic Operad. Derivations on free operads are studied in Section 6.3.3, and in particular Proposition 6.3.6 states that a derivation $\mathrm{Free}(F) \to \mathrm{Free}(F)$ is the same thing as a map $F \to \mathrm{Free}(F)$.
This is equivalent to requiring that for homogenous $f,g$, one has: $$d(f \circ_i g) = d(f) \circ_i g + (-1)^{|f|} f \circ_i d(g).$$
Note that derivations are rarely morphisms of operads. It follows from the axioms that if $u \in P(1)$ is the unit of the operad, then $d(u) = 0$. However, a morphism of operads has to satisfy $d(u) = u$. So, a derivation is a morphism of operads if and only if $u = 0$, in which case $P = 0$ is the zero operad.