If $F(x,y)=\int_{f(x,y)}^{g(x,y)}h(x,y,t)\, dt$, then find $\partial F/\partial x$. 
Let $f,g:\mathbb{R}^2\to \mathbb{R}$ and $h:\mathbb{R}^3\to\mathbb{R}$ be $C^1$, define $F(x,y)=\int_{f(x,y)}^{g(x,y)}h(x,y,t)\, dt$. Find $\frac{\partial F}{\partial x}$.

So I split $$F(x,y)=\int_0^{g(x,y)}h(x,y,t)\, dt-\int_0^{f(x,y)}h(x,y,t)\, dt.$$
And then using FTC I believe I have $$F(x,y)=H(x,y,g(x,y))-H(x,y,0)-H(x,y,f(x,y))+H(x,y,0)=H(x,y,g(x,y))-H(x,y,f(x,y))$$
Then differentiating this new thing with respect to x.
$$\frac{\partial F}{\partial x}=\frac{\partial H}{\partial x}+\frac{\partial H}{\partial y}\frac{\partial y}{\partial x}+\frac{\partial H}{\partial g(x,y)}\frac{\partial g(x,y)}{\partial x}-\frac{\partial H}{\partial x}-\frac{\partial H}{\partial y}\frac{\partial y}{\partial x}-\frac{\partial H}{\partial f(x,y)}\frac{\partial f(x,y)}{\partial x}$$
Is this correct? From here I would simplify the expression and expand out the partials. But I'm not sure that the FTC works in this way for $H$, since it's a function of 3 variables.
 A: Your approach is correct.
Since $h : \mathbb{R}^3 \to \mathbb{R}$ is $C^1$ and hence continuous, for each fixed $x,y \in \mathbb{R}$ the function $t \mapsto h(x,y,t)$ has an antiderivative of the form $H(x,y,t)$, and we indeed have that
$$
\int_{f(x,y)}^{g(x,y)} h(x,y,t)\, dt = H(x,y,g(x,y)) - H(x,y,f(x,y)).
$$
The chain rule you have used for computing the partial derivative $\partial H / \partial x$ is also accurate. However it can be simplified by noting (as @RafaBudria mentions in the comments) that $y$ is independent of $x$, so the terms containing $\partial y / \partial x$ can be removed as $\partial y / \partial x = 0$. So your final answer should be
$$
\frac{\partial F}{\partial x} = \frac{\partial H}{\partial g} \frac{\partial g}{\partial x} - \frac{\partial H}{\partial f} \frac{\partial f}{\partial x}.
$$
Lastly, I am not sure what the usual notation for the partial derivatives is, but perhaps writing $\partial H / \partial t$ in place of $\partial H / \partial f$ and $\partial H / \partial g$ would be an improvement.
