Time derivative of a line integral Question:
What is the derivative of 
$$ \frac{d}{dt} \int_{(x_0,y_0,z_0)}^{(x,y,z)} \vec{F}(x,y,z,t) \cdot d\vec{r}$$
Now in order to compute any single integral, I need to give you a path. So far, all we know is that we are integrating $\vec{F}$ from the point $(x_0, y_0, z_0)$ to the point $(x, y, z)$ along some path. Let me specify that path. What is the derivative of
$$ \frac{d}{dt} \int_{t_0}^{t} \vec{F}(c(t),t) \cdot \vec{c}\;'(t)dt $$
where $c(t) = (x(t), y(t), z(t))$ is the path and $\vec{c}\;'(t) = (x'(t), y'(t), z'(t))$ is the tangent vector to the path. 
Thoughts: Derivatives require a path. You can't take a derivative unless you know what path you are taking the derivative over. Likewise, integrals/antiderivatives require a path. You can't create an antiderivative unless you know what path you are doing the integral over. If I created the antiderivative
$$\int_{t_0}^{t} \vec{F}(c(t)) \cdot \vec{c}\;'(t)dt $$
This would be the antiderivative over the path $c(t)$. Therefore if I take the derivative $d/dt$, this indicates the derivative over the same path $c(t)$. And since we know derivatives and antiderivatives are inverse operations (over the same path),
$$ \frac{d}{dt} \int_{t_0}^{t} \vec{F}(c(t)) \cdot \vec{c}\;'(t)dt = \vec{F}(c(t)) \cdot \vec{c}\;'(t)$$
That's what I think at least. My problem is that my integrand involves an extra $t$ (I'm asking this because in order to understand time-dependent forces and energy consequences in physics, I'd like to understand derivatives of line integrals).
Useful break down of my question:
These would be useful to solve. Just to make the problem look simpler, what is
$$\frac{d}{dt} \int_{t_0}^{t} f(x(t),y(t))x'(t)dt $$ This is the antiderivative over the path $x(t)$. But the derivative $d/dt$ is with respect to the path $(x(t), y(t))$. So would this become, using the chain rule
$$ \frac{\partial }{\partial x} \Bigg [\int_{t_0}^{t} f(x(t),y(t))x'(t)dt\Bigg]\frac{dx}{dt} + \frac{\partial}{\partial y}\Bigg[\int_{t_0}^{t} f(x(t),y(t))x'(t)dt\Bigg]\frac{dy}{dt}$$
Is the second bracket zero? Another useful, simpler looking, problem:
$$\frac{d}{dt}\int_{t_0}^t f(x(t),t)x'(t)dt $$
Which might become:
$$ \frac{\partial }{\partial x} \Bigg [\int_{t_0}^{t} f(x(t), t)x'(t)dt\Bigg]\frac{dx}{dt} + \frac{\partial}{\partial t}\Bigg[\int_{t_0}^{t} f(x(t), t)x'(t)dt\Bigg]\frac{dt}{dt}$$
Is the first bracket just $f(x(t),t)$? If I write this first bracket as 
$$ \int_{x_0}^x f(x, t)dx$$ Then a $\partial/\partial x$ of this antiderivative should return $f(x,t)$. However I'm getting confused by the abstraction of everything and the actual computation. This last integral seems to imply that the path is just the $x$-axis. But the real path is $x(t)$. You need to do the integral with all $t$'s then. But I still feel like
$$ \frac{\partial }{\partial x}\int_{t_0}^{t} f(x(t), t) x'(t)dt = f(x(t),t) ?$$
If you read all of this, thank you. Any help would be greatly appreciated
 A: You don’t always need the path. That depends on the nature of the force. If it is a conservative force, then the integral is path independent. In other words, if $\vec{F}=\nabla \varphi$ in some domain for some $\varphi$, then $$\int_{c_1} \vec{F} = \int_{c_2} \vec{F}$$ where $c_1,c_2$ are paths in that domain. (Easily proved by applying Stokes’ theorem).
An example is if the sole force acting on a body is gravity.  The work done (and thus the change in energy) between points A and B can be computed by just knowing the points A and B. It doesn’t matter which path was taken. 
So then the integral and its derivative can be taken without knowing the path. In your case, do you know anything like that about the force? 
A: So you have a time dependent vector field $\vec{F}(x,y,z,t)$, and for a curve $\vec{c}(s)$ in $\mathbb{R}^3$ you're interested in the integral of $\vec{F}$  along $\vec{c}$ from $s=t_0$ to $s=\hat{t}$. Notice I'm using a different symbol for the argument of $c$; $c$ maps into the domain of $\vec{F}$ so strictly speaking you should use a different symbol for its argument if you're going to compose the functions. Also, different symbols must be used for the limits of integration. Now, you're interested in the derivative:
$$\frac{d}{d\hat{t}}\int_{t_0}^\hat{t}\vec{F}(\vec{c}(s),s)\cdot\vec{c}'(s)ds$$
But this is just the fundamental theorem of calculus from single variable calculus: the expression inside the integral is a single variable function of $s$ (a dot product gives you a real number), which we are integrating, and then differentiating with respect to one of the limits of integration. All we need to know is that the integrand meets the criterion required by the FTC. Since this is physics, $\vec{F}$ and $\vec{c}$ are probably differentiable or at least continuous, and the dot product of two continuous functions is again continuous, which makes the integrand continuous, which is good enough to make the FTC work. The result is
$$\frac{d}{d\hat{t}}\int_{t_0}^\hat{t}\vec{F}(\vec{c}(s),s)\cdot\vec{c}'(s)ds=\vec{F}(\vec{c}(\hat{t}),\hat{t})\cdot\vec{c}'(\hat{t})$$
