How do I differentiate under the line integral along an open curve? Is there a way to perform the following differentiation under the line integral sign ?
$$ I = \frac{d}{dt}\left(\int_{\Gamma\left(t\right)}\vec{u}\cdot d\vec{x}\right) $$
provided the vector $ \vec{u}=\vec{u}\left(\vec{x},t\right)$  varies in space and also varies with the variable $ t $. And provided that $ \vec{u} $ may not be conservative.
The curve $\Gamma$ may also be allowed to depend on the variable $ t $ and may even be allowed to undergo translations and rotations. But, the most important thing about $\Gamma$ is that it be an open curve and not a closed one. This is because I know that the closed curve would be a special case of the 2-D Liebnitz rule.
Is there a general formula for this operation ? and, if so, how could one prove it?

It seems to be possible to consider the problem i a much more general form.
If the vector $\vec{v_{n}}$ is the normal velocity field of all points along the curve $\Gamma$ and if the vector $\vec{v_{t}}$ is the tangent velocity field along the curve $\Gamma$, then this argument concludes that the derivative under the line integral should be
$$\frac{d}{dt}\int_{\Gamma\left(t\right)}\vec{u}\cdot d\vec{x}=\left[\vec{u}\left(\vec{x_{2}},t\right)\cdot \vec{v_{t_2}} -\vec{u}\left(\vec{x_{1}},t\right)\cdot \vec{v_{t_1}}\right]+\int_{\Gamma\left(t\right)}\left[\frac{\partial \vec{u}}{\partial t}+\left(\vec{v_{n}}\cdot\nabla\right)\vec{u}\right]\cdot d\vec{x} $$
$$+\int_{\Gamma\left(t\right)}\vec{u}\cdot\frac{\partial \vec{v_{n}}}{\partial l}dl$$
where $\vec{x_{1}}$ and $\vec{x_{2}}$ are the coordinates of the initial and final points on $\Gamma$ respectively, while $\vec{v_{t_1}}$ and $\vec{v_{t_2}}$ are the corresponding tangential velocity components. The length along the curve $\Gamma$ i denoted by $l$.
