Differentiating the absolutely continuous invariant measure of an   interval map f with respect to f
  Let the map $f:[-1,1]\to[-1,1]$ have a.c.i.m. $\rho$ (absolutely continuous $f$-invariant measure with respect to Lebesgue). Let $\delta\rho$ be the change of $\rho$ corresponding to a perturbation $X=\delta f\circ f^{-1}$ of $f$. Formally we have, for differentiable $A$, $$ \delta\rho(A)=\sum_{n=0}^\infty\int\rho(dx) X(x){d\over dx}A(f^nx) $$ but this expression does not converge in general. For $f$ real-analytic and Markovian in the sense of covering $(-1,1)$ $m$ times, and assuming an {\it analytic expanding} condition, we show that $$\lambda\mapsto\Psi(\lambda)=\sum_{n=0}^\infty\lambda^n \int\rho(dx) X(x){d\over dx}A(f^nx) $$ is meromorphic in ${\bf C}$, and has no pole at $\lambda=1$. We can thus formally write $\delta\rho(A)=\Psi(1)$. 
David Ruelle