thermodynamics - Is $delta Q – delta W$ a state function?

I know that internal energy, $Q+W$ is a state function. But $$dU=\delta Q-\delta W,$$
is the change in internal energy, where $dU$ is change in internal energy, $\delta Q$ is the heat supplied to the system and $\delta W$ is the work done by the system. Is this a state function too?

Answer

The quantity $dU=\delta Q-\delta W$ is not a state function simply because it is not a function, it is a differential of a function. In that case a differential of the state function $U$.

The fact that there is a function $U$ such that the differential $\delta Q-\delta W$ is the differential of $U$ means that the differential is exact or $U$ is a state function, i.e., it has a definite value for every point in the space of configuration. In another words, the change in energy depends only on the initial and final states.