Le $M$ be a Riemannnian manifold. Is possible to express the divergence of an $(1,1)$ tensor $T$, i. e., an operator by a general formula as happen for the Ricci tensor where we have the well known second contracted Bianchi identity ${\rm div}(Ric)=\frac{1}{2}{\rm d}R$. If the answer is not in general, would it be true if $T=\nabla^2 f$, the Hessian of a smooth function $f$?
Summarizing: How to calculate ${\rm div}(T)$ or at least ${\rm div}(\nabla^2 f)$ explicitly in terms of ${\rm trace}(T)$ or some other known quantity?
In a course about general relativity, we defined the divergence of a (generic) tensor as
$div \ T \equiv tr_1^1 (\nabla T)$ ;
is this what you are looking for?