The Lie Bracket of a Poisson Algebroid

Question

Given a Poisson manifold $(M,\pi)$, define a bracket on $\Omega^1(M)$ by $$ [\alpha,\beta]=\mathcal{L}_{\pi^\sharp(\beta)}\alpha-\mathcal{L}_{\pi^\sharp(\alpha)}\beta-d\pi(\alpha,\beta) $$ Anti-symmetry is clear. But we need the Leibniz rule $[\alpha,f\beta]=f[\alpha,\beta]+(\rho(\alpha)f)\beta$, where the anchor map is $\rho(\alpha)=\pi(\cdot,\alpha)$. This gives me some trouble. Each time I try to prove this, I get the following term: $$d\pi(\alpha,f\beta)=d(f\pi(\alpha,\beta))=\pi(\alpha,\beta)df+fd\pi(\alpha,\beta)$$ The $df$ term should not be there. But I do not see how to make it go away, given the Leibniz rule of the exterior derivative. Could someone tell me how this works?