Partial Derivative of Line Integral as a Potential of F Context to the question:
Say $ \{F_{k} \} \to F$ uniformly on a compact subset $K \subset T$, for $ \{F_{k} \}$ a sequence of conservative vector fields and $T$ open and connected.
I've shown that the limit $F$ must itself be conservative.
Suppose we define $\gamma_{x}$ to be a straight curve between a point $a \in T$ and $x \in K$. Then letting $f_{k}(x) = \int_{\gamma_{x}}F_{k} dl$ it can be shown that $f_{k}(x)$ converges uniformly to $ f(x) = \int_{\gamma_{x}}F dl$ on the compact subset.
I want to show that $\bigtriangledown f(x) = F$:
for that I've tried $$\lim_{h\to 0} \frac{f(x_{1},..x_{i}+h,..x_{n})-f(x_{1},..x_{i},..x_{n})}{h} = \lim_{h\to 0}  \int_0^1 F^{i}((1-t)a+tx_{i,h})dt$$
for $x_{i,h} = (x_{1},..x_{i}+h,..x_{n})$
but can't see why $\frac {\partial f}{\partial x_{i}}(x) = F^{i}$ (coordinate $ i$)
Thanks for your help!
edit* I fixed the value of $\frac{\partial f}{\partial x_{i}}(x)$ the result now almost follows from the mean value theorem for integrals, but for $ t \in [0,1]$, do you know what is wrong?
 A: $\newcommand{\dd}{\partial}$Here's an outline:


*

*The function $f_{k}$ is a potential for the conservative field $F_{k}$. The usual trick for this is, for each $i = 1$, ..., $n$ in turn, to replace the path $\gamma_{x}$ with a path $\gamma_{x}^{i}$ from $a$ to $x$ that arrives at $x$ along a unit-speed segment parallel to the $x_{i}$-axis, i.e., that satisfies $\gamma_{x}^{i}(1 + h) = x + h\mathbf{e}_{i}$ for $|h|$ small. The resulting integral agrees with $f_{k}$ because $F_{k}$ is conservative (has path-independent line integrals), and we have
\begin{align*}
\frac{f_{k}(x + h\mathbf{e}_{i}) - f_{k}(x)}{h}
  &= \frac{1}{h} \int_{0}^{h} F_{k}(x + t\mathbf{e}_{i}) \cdot \mathbf{e}_{i}\, dt \\
  &= \frac{1}{h} \int_{0}^{h} F_{k}^{i}(x + t\mathbf{e}_{i})\, dt.
\end{align*}
As $h \to 0$, the right side approaches $F^{i}(x)$ by continuity of $F^{i}$. That is,
$$
\frac{\dd f_{k}}{\dd x_{i}} = F_{k}^{i}.
$$
(It's possible this is the entire hint you're looking for.)

*The sequence $(f_{k})$ converges uniformly on compacta to the function
$$
f(x) = \int_{\gamma_{x}} \lim_{k \to \infty} F_{k}\, dl = \int_{\gamma_{x}} F\, dl.
$$

*By the theorem mentioned in the comment, whose proof amounts to the fundamental theorem and the fact that uniform limits commute with integrals,
$$
\nabla f = \lim_{k \to \infty} \nabla f_{k}
  = \lim_{k \to \infty} F_{k}
  = F.
$$
In other words, run through the gradient computation in 1. above, commuting the uniform limit with integration.
