Why there is added a partial time derivative in formula for time derivative of potential energy? [duplicate]

In proving the total energy in conservative field is constant we have this equation(picture) why it added partial derivative? Why? I mean where it did come from? marked as duplicate by Kyle Kanos, Colin McFaul, BMS, Brandon Enright, Kyle OmanJul 9 '14 at 16:37

The function $U = U(x_1, x_2, \dots x_k, t)$ would be an example of a potential energy function explicitly dependent on time. In your case, you have the function $U = U(x_1, x_2,\dots x_k)$, where it is understood that for each $x_i,\, i \in \left\{1,\dots k\right\}$, we have an implicit dependence $x_i = x_i(t)$. The total derivative of $U$ is $$dU = \sum_i \frac{\partial U}{\partial x_i}d x_i + \frac{\partial U}{\partial t}dt$$ and furthermore, $$\frac{dU}{dt} = \sum_i \frac{\partial U}{\partial x_i}\frac{d x_i}{dt} + \frac{\partial U}{\partial t}.$$
Lets make up an example. $$U(x,y,z,t) = E_0\left(x^2+y^2+\alpha\,z^4 -\beta\,t\,z^2\right)$$ In that case parital derivatives are: $$\frac{\partial U}{\partial x} = 2xE_0,\; \frac{\partial U}{\partial x} = 2yE_0,\; \frac{\partial U}{\partial z} = \left(4\alpha z^3 -2\beta z\right)E_0,\; \frac{\partial U}{\partial t} = -\beta\,z^2E_0$$ Now you have a particle that moves like: $$x(t) = R\cos \omega t,\quad y(t) = R\sin \omega t,\quad z(t) = vt$$ If we substitute these, then we can have $U$ as a function of $t$ only: $$U(t) = U\left(x(t),y(t),z(t),t\right) = E_0\left(R^2 + \alpha\,v^4t^4 -\beta v^2t^3\right)$$ And the full time derivative is: $$\frac{dU}{dt} = \left(4\alpha\,v^4t^3-3\beta v^2t^2\right)E_0$$ I'm leaving it for you to check that you'll have the same result if you use the chain rule together with partial time derivative $$\frac{dU}{dt} = \frac{\partial U}{\partial t} + \frac{\partial U}{\partial x}\frac{dx}{dt} + \frac{\partial U}{\partial y}\frac{dy}{dt} +\frac{\partial U}{\partial z}\frac{dz}{dt}$$