Poisson Bracket of a Quantity Involving a Differential I am working through Warren Siegel's "Fields" and have come across the following exercise on p. 58 involving an action measure and a symmetry generator:

Exercise IA4.1.
For general variables ($q^m$,$p_m$) and generator $G$, show from the definition of the Poisson bracket that $\delta(dq^m p_m) = -d\left(G - p_m \frac{\partial G}{\partial p_m}\right)$ and that this vanishes for any coordinate transformation.

As defined, the $\delta$ of a quantity $A$ means the Poisson bracket $\{A,G\}$. Given the definition of a Poisson bracket I'm familiar with,
$$\{A(q,p), B(q,p)\} = \frac{\partial A}{\partial q^m}\frac{\partial B}{\partial p_m} - \frac{\partial A}{\partial p^m}\frac{\partial B}{\partial q_m}$$
I am unsure how to handle the differential appearing in $\delta(dq^m p_m)$. What is the correct way to apply a Poisson bracket to such a quantity?
 A: It appears that the question requires a bit more nuance. Rather than defining the $\delta$ as the direct application of a Poisson bracket, one can think of $\delta (d q^m p_m)$ as  
$$ \delta (d q^m p_m) = (dq^m + \delta (dq^m))(p_m + \delta p_m) - dq^m p_m $$
Now $\delta (dq^m) = d (\delta q^m)$ and one finds (using the Poisson bracket) that
$$ \delta q^m = \{q^m, G\} = \frac{\partial G}{\partial p_m}$$
$$ \delta p_m = \{p_m, G\} = -\frac{\partial G}{\partial q^m}$$
Thus, omitting terms $2^{\text{nd}}$ order in $\delta$s, 
$$ \delta (d q^m p_m) = \delta(dq^m)p_m + dq^m \delta p_m = p_m d \left( \frac{\partial G}{\partial p_m} \right) - dq^m \frac{\partial G}{\partial q^m}$$
whereas the RHS of the question is
$$ -d(G - p_m\frac{\partial G}{\partial p_m}) = p_m d\left(\frac{\partial G}{\partial p_m} \right) + dp_m \frac{\partial G}{\partial p_m} - dG = p_m d \left( \frac{\partial G}{\partial p_m} \right) - dq^m \frac{\partial G}{\partial q^m}$$
Thus we see that the LHS and RHS are equal.
