Let $(X, \mathcal{X})$ and $(Y, \mathcal{Y})$ be a two measurable spaces and $f: X \to Y$ be a measurable map.
Suppose that $Q$ and $\mu$ are sigma-finite measures on $X$ such that the absolute continuity condition $Q \ll \mu$ holds, that is $\mu(A) = 0 \implies Q(A) = 0$ for all $A \in \mathcal{X}$.
Then $Q$ has a Radon-Nikodym derivative (density) $\displaystyle \frac{\mathrm{d}Q}{\mathrm{d}\mu}(x)$ with respect to $\mu$.
Define $Q \circ f^{-1}$ to be the pushforward measure of $Q$ with respect to $f$, defined by the rule $(Q \circ f^{-1})(B) = Q(f^{-1}(B))$ for all $B \in \mathcal{Y}$. Define $\mu \circ f^{-1}$ analogously.
It can be seen that $(Q \circ f^{-1}) \ll (\mu \circ f^{-1})$, since $$(\mu \circ f^{-1})(B) = 0 \implies \mu(f^{-1}(B))=0 \implies Q(f^{-1}(B))=0 \implies (Q \circ f^{-1})(B) = 0,$$
so that $(Q \circ f^{-1})$ has a density $\displaystyle \frac{\mathrm{d}(Q \circ f^{-1})}{\mathrm{d}(\mu \circ f^{-1} )}(y)$ with respect to $(\mu \circ f^{-1})$.
What can be said about the density $\displaystyle \frac{\mathrm{d}(Q \circ f^{-1})}{\mathrm{d}(\mu \circ f^{-1} )}(y)$, for general $f$?
What is its relationship to $\displaystyle \frac{\mathrm{d}Q}{\mathrm{d}\mu}(x)$?
How about if $f$ is a bijection?