First law of thermodynamics as conservation of energy We have that $\Delta U = Q + W$. What I don't see is how this formula relates to the law of conservation of energy. Can someone please clarify?
Does this mean that $\frac{dU}{dt}=\frac{dQ}{dt}+\frac{dW}{dt}=0$, so that $\frac{dQ}{dt}=-\frac{dW}{dt}$?
 A: Since your second question has been answered in the comments, I will answer your first question.
Let's examine in words what $\Delta U=Q+W$ means: "Any change in internal energy arises from a flow of heat into/out of the system and/or work done by/on the system." Put differently: "The only two ways internal energy can change are if heat flow occurs or if work is done." Both heat flow and work are examples of energy transfers (indeed, both have units of energy). Also, note that any energy transfer to the system that is not classified as heat flow is automatically classified as work done on the system.
Putting this all together, we can restate the equation as follows: "In order to change the internal energy of a system, you must add or subtract energy from the system." This is an indirect way of stating that energy is conserved.
It might be easier to see this in the case of an isolated system, where no external heat flow or work occurs, so $Q=W=0$. In that case, the law reads: "In an isolated system, total internal energy does not change." This is a direct statement of conservation of energy.
A: The equation of the first law of Thermodynamics does not directly imply the conservation of energy; rather the first law of Thermodynamics is a consequence of the conservation of energy when applied to Thermodynamics( systems involving heat, temperature etc.)
The conservation of energy was known somewhat by scientists like Galileo in 1638 (knew about potential and kinetic energy conservation in a pendulum) and others before him too. This was before 1850 when Rudolf Clausius and Lord Kelvin stated the First Law of Thermodynamics for energy conservation in systems involving heat energy.
To answer your second question, yes if you take time derivatives on both sides that equation results but that does not mean that the rate of change of internal energy($U$) is zero, it's dependent on how the other factors like rate of heat energy supplied($\frac{dQ}{dt}$) or rate of work done on the system($\frac{dW}{dt}$) changes with time. If both of these are zero or they are equal in magnitude and opposite in sign then only can the rate of change of $U$ be zero. It's a special condition when internal energy does not increase not a condition valid for all times. Ofcourse the conservation of energy can hold even if there is a change in $U$, as then the energy comes from the surroundings to add to the system.
A: 
We have that $\Delta U=Q+W$. What I don't see is how this formula relates to the law of conservation of energy. Can someone please clarify?

The law of energy conservation says that no energy can ever appear out of the blue or disappear into thin air:
$$E_{1}=E_{2}$$
where $_1$ is before and $_2$ after a certain point in time. Let's plug in all the energies that we can think of:
$$K_{trans,1}+K_{rot,1}+K_{thermal,1}+U_{grav,1}+U_{elast,1}+U_{electr,1}+U_{chem,1}+\cdots=\\K_{trans,2}+K_{rot,2}+K_{thermal,2}+U_{grav,2}+U_{elast,2}+U_{electr,2}+U_{chem,2}+\cdots
$$
$K$ and $U$ are different types of kinetic and potential energies. If you add energy by heating up $Q$ or doing work $W$ on the system, then you just put that in as well$$K_{trans,1}+K_{rot,1}+K_{thermal,1}+U_{grav,1}+U_{elast,1}+U_{electr,1}+U_{chem,1}+\cdots+Q+W=\\K_{trans,2}+K_{rot,2}+K_{thermal,2}+U_{grav,2}+U_{elast,2}+U_{electr,2}+U_{chem,2}+\cdots
$$
And now all we need is to reduce all this. In thermodynamic systems you might want to skip the extrinsic kinetic energies and the gravitational potentials are often negligible:
$$\require{cancel}
\cancel{K_{trans,1}}+\cancel{K_{rot,1}}+K_{thermal,1}+\cancel{U_{grav,1}}+U_{elast,1}+U_{electr,1}+U_{chem,1}+\cdots+Q+W=\\\cancel{K_{trans,2}}+\cancel{K_{rot,2}}+K_{thermal,2}+\cancel{U_{grav,2}}+U_{elast,2}+U_{electr,2}+U_{chem,2}+\cdots
$$
And all the rest can just be collected in one symbol $U_{internal}$ (the $U$ is often used, but it does of course not mean potential energy now but rather internal energy):
$$
U_{internal,1}+Q+W=U_{internal,2}\\
\Delta U_{internal}=Q+W
$$
Exactly how I cancelled out the ignored energies and exactly which are included as internal energy can be up for discussion. But these steps show, how the formula from the 1st law of thermodynamics arrive directly from energy conservation - it is the energy conservation law, just for a thermodynamic system.

Does this mean that $\frac{dU}{dt}=\frac{dQ}{dt}+\frac{dW}{dt}=0$, so that $\frac{dQ}{dt}=−\frac{dW}{dt}$?

No. It doesn't mean that. And I don't really see why you would think so. The rate of change in internal energy $\frac{dU}{dt}$ can easily be non-zero, it just has to equal the rate that energy is added either as heat $\frac{dQ}{dt}$ or work $\frac{dW}{dt}$.


*

*For example: If I add 2 Joules of heat every second and also add 1 Joule of energy as work every second, then the gain in internal energy every second is $\frac{dU}{dt}=3\;\mathrm J$.


If $\frac{dU}{dt}$ happens to be zero in a specific case, then it just means that any heat added is taken out as work equally, or the other way around, and none is "kept" inside the system/object. This is the case for a car engine, e.g. After having heated up, it doesn't go any warmer even though it continues to drive. There is a balance between cooling and heating and the work performed. This is called a cyclic process.
A: (I will adopt the sign convention $\Delta U =Q - W$ because I like it more)
If you were able to give a complete microscopic description of the system you are considering, then you would simply write 
$$\Delta U = -W =  -\sum_{i=1}^N \int_{C_i} \mathbf  F_i \{\mathbf r_1, ..., \mathbf r_N\}\cdot d \mathbf r_i$$
Indeed, from a microscopic point of view all the forces of interest in most problems we consider are conservative, so we just have to apply the usual energy-work relation from mechanics.
When we do thermodynamics, we are renouncing to give a complete microscopic description of the system. For example, in an hydrostatic system we will usually consider the macroscopic work done by the system, i.e. $W = -P \Delta V$. 
In doing this, we will notice that there $\Delta U = - W$ is not valid anymore, and that there is some missing energy: this missing energy is what we call heat ($Q$), and it represent the microscopic work that we were not able to include in our description of the system.
So, to restore the mechanical conservation of energy $\Delta U = -W$, we have to put by hand this term $Q$ in, obtaining the well known thermodynamic statement of conservation of energy.
This is why $\Delta U = Q - W$ is a statement of conservation of energy: the energy that is apparently lost when we go from a microscopic to a macroscopic description is the microscopic work, which we call heat.
Now you ask why we don't have $dU/dt = 0$, or equivalently why we don't have $\Delta U=0$ in general: this is because if the system is allowed to exchange heat and work with the environment, then it can lose or gain energy: think for example of a gas expanding, or a piece of metal cooling.
However, if we consider an isolated system, i.e. a system that cannot exchange heat nor work with the environment, then $Q=W=0$, so that $\Delta U=0$. 
From here, we could take -I admit it- quite a huge leap and say this: since the Universe is (supposedly) an isolated system, then, for the whole universe, we will always have $\Delta U=0$, i.e. on a universal scale, energy is always conserved: it can move around in the form of work and heat (a.k.a. microscopic work), but it can never disappear or appear from nowhere*.

* Except for very short amounts of time, at least.
