How can I compute the partial derivative under the line integral sign if the integrand is non-conservative?

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Consider the following integral

$$I=\int_C^{\vec{y}_{0},\vec{y}} \vec{F}(\vec{x})\cdot d\vec{x}$$

where, for simplicity, $\vec{y}_{0}$ is some constant initial point on $C$ (which is an explicitly known curve) and $\vec{y}$ is some variable end point on $C$ and it is known that

$$\oint_{C_0} \vec{F}(\vec{x})\cdot d\vec{x} \neq 0 $$

for any closed curve $C_{0}$.

Does it still make sense to speak of the gradient

$$\left(\frac{\partial}{\partial y_{1}},\frac{\partial}{\partial y_{2}},\frac{\partial}{\partial y_{3}}\right)I$$

of the integral with respect to $\vec{y} = \left(y_{1},y_{2},y_{3}\right)$ ?

If so, then three questions follow.

1) If this gradient exists then doesn't that contradict the premise that $\vec{F}$ is non-conservative?

2) Is there an equivalent of the Leibniz integral rule which specifies such a gradient?

3) How does it depend of the curve $C$?

Otherwise, how can we characterize the variational properties of this differentiation under the integral?

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$I$, as you have given it, is not a well-defined function of $y$. There may be two different curves ending at $y$, yielding two different line integrals. Which one should I choose to compute the value of $I$ at any given $y$?

Only when the vector field is conservative do we know that this function is well-defined. Only then may we take its gradient.